Prove logarithmic inequality:












1












$begingroup$


Prove $${log_{2016}}^{2016}sqrt{1+frac{a_2}{a_1}}quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_3}{a_2}}quad+quad...quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_1}{a_{2016}}}quad ge frac{1}{5+{log_2}63}$$



(It's 2016th root, not $log_{2016}$ to the power of 2016! I didn't know how to write 2016th root).



I have a solution, but I don't understand it. It says in the solution that the equation above is equivalent to
$$log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}} ge log_{2016}2^{2016}$$



I can get the left side to
$$frac1{2016}log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}}$$



But I don't know how to get the right side as in the solution. How do I get that result?






algebra-precalculus







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Prove $${log_{2016}}^{2016}sqrt{1+frac{a_2}{a_1}}quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_3}{a_2}}quad+quad...quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_1}{a_{2016}}}quad ge frac{1}{5+{log_2}63}$$



    (It's 2016th root, not $log_{2016}$ to the power of 2016! I didn't know how to write 2016th root).



    I have a solution, but I don't understand it. It says in the solution that the equation above is equivalent to
    $$log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}} ge log_{2016}2^{2016}$$



    I can get the left side to
    $$frac1{2016}log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}}$$



    But I don't know how to get the right side as in the solution. How do I get that result?






    algebra-precalculus







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Prove $${log_{2016}}^{2016}sqrt{1+frac{a_2}{a_1}}quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_3}{a_2}}quad+quad...quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_1}{a_{2016}}}quad ge frac{1}{5+{log_2}63}$$



      (It's 2016th root, not $log_{2016}$ to the power of 2016! I didn't know how to write 2016th root).



      I have a solution, but I don't understand it. It says in the solution that the equation above is equivalent to
      $$log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}} ge log_{2016}2^{2016}$$



      I can get the left side to
      $$frac1{2016}log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}}$$



      But I don't know how to get the right side as in the solution. How do I get that result?






      algebra-precalculus







      share|cite|improve this question









      $endgroup$




      Prove $${log_{2016}}^{2016}sqrt{1+frac{a_2}{a_1}}quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_3}{a_2}}quad+quad...quad+quad {log_{2016}}^{2016}sqrt{1+frac{a_1}{a_{2016}}}quad ge frac{1}{5+{log_2}63}$$



      (It's 2016th root, not $log_{2016}$ to the power of 2016! I didn't know how to write 2016th root).



      I have a solution, but I don't understand it. It says in the solution that the equation above is equivalent to
      $$log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}} ge log_{2016}2^{2016}$$



      I can get the left side to
      $$frac1{2016}log_{2016}frac{(a_1+a_2)(a_2+a_3)...(a_{2016}+a_1)}{a_1a_2...a_{2016}}$$



      But I don't know how to get the right side as in the solution. How do I get that result?






      algebra-precalculus




      algebra-precalculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 22:35









      PeroPero

      1277




      1277






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          We have that



          $$frac{1}{5+{log_2}63}=frac{1}{log_2 32+{log_2}63}=frac1{log_2 (32cdot 63)}=frac1{log_2 2016}=log_{2016}2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
            $endgroup$
            – Pero
            Dec 10 '18 at 22:58












          • $begingroup$
            Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:05










          • $begingroup$
            We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:11












          • $begingroup$
            @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:13












          • $begingroup$
            Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:19











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          1 Answer
          1






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          1 Answer
          1






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          active

          oldest

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          1












          $begingroup$

          We have that



          $$frac{1}{5+{log_2}63}=frac{1}{log_2 32+{log_2}63}=frac1{log_2 (32cdot 63)}=frac1{log_2 2016}=log_{2016}2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
            $endgroup$
            – Pero
            Dec 10 '18 at 22:58












          • $begingroup$
            Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:05










          • $begingroup$
            We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:11












          • $begingroup$
            @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:13












          • $begingroup$
            Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:19
















          1












          $begingroup$

          We have that



          $$frac{1}{5+{log_2}63}=frac{1}{log_2 32+{log_2}63}=frac1{log_2 (32cdot 63)}=frac1{log_2 2016}=log_{2016}2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
            $endgroup$
            – Pero
            Dec 10 '18 at 22:58












          • $begingroup$
            Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:05










          • $begingroup$
            We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:11












          • $begingroup$
            @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:13












          • $begingroup$
            Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:19














          1












          1








          1





          $begingroup$

          We have that



          $$frac{1}{5+{log_2}63}=frac{1}{log_2 32+{log_2}63}=frac1{log_2 (32cdot 63)}=frac1{log_2 2016}=log_{2016}2$$






          share|cite|improve this answer









          $endgroup$



          We have that



          $$frac{1}{5+{log_2}63}=frac{1}{log_2 32+{log_2}63}=frac1{log_2 (32cdot 63)}=frac1{log_2 2016}=log_{2016}2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 22:38









          gimusigimusi

          92.8k84494




          92.8k84494












          • $begingroup$
            Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
            $endgroup$
            – Pero
            Dec 10 '18 at 22:58












          • $begingroup$
            Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:05










          • $begingroup$
            We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:11












          • $begingroup$
            @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:13












          • $begingroup$
            Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:19


















          • $begingroup$
            Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
            $endgroup$
            – Pero
            Dec 10 '18 at 22:58












          • $begingroup$
            Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:05










          • $begingroup$
            We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:11












          • $begingroup$
            @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
            $endgroup$
            – gimusi
            Dec 10 '18 at 23:13












          • $begingroup$
            Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
            $endgroup$
            – Pero
            Dec 10 '18 at 23:19
















          $begingroup$
          Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
          $endgroup$
          – Pero
          Dec 10 '18 at 22:58






          $begingroup$
          Oh I see :). After we get this we can prove that the new inequality is true using $A ge G$. In the solution it says that this inequality is true if and only if $a_1=a_2=...=a_{2016}$. Why is it not true for any value of $a_1, a_2,...,a_{2016}$?
          $endgroup$
          – Pero
          Dec 10 '18 at 22:58














          $begingroup$
          Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
          $endgroup$
          – gimusi
          Dec 10 '18 at 23:05




          $begingroup$
          Could you add your derivation by AM-GM? (I suppose you are referrign to that with $Age G$)
          $endgroup$
          – gimusi
          Dec 10 '18 at 23:05












          $begingroup$
          We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
          $endgroup$
          – Pero
          Dec 10 '18 at 23:11






          $begingroup$
          We can rewrite the second equation in my question above as (if we remove $log_{2016}$): $(a_1+a_2)...(a_{2016}+a_1) ge 2^{2016} cdot a_1...a_{2016}$, and this can be easily proven by solving AM-GM for $a_1,a_2$; then for $a_2,a_3$; and so on and then multiplying.
          $endgroup$
          – Pero
          Dec 10 '18 at 23:11














          $begingroup$
          @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
          $endgroup$
          – gimusi
          Dec 10 '18 at 23:13






          $begingroup$
          @Pero Then I think it suffices to recall that equality holds for AM-GM when $a_1=a_2=...=a_{2016}$. Its just a property of AM-GM.
          $endgroup$
          – gimusi
          Dec 10 '18 at 23:13














          $begingroup$
          Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
          $endgroup$
          – Pero
          Dec 10 '18 at 23:19




          $begingroup$
          Yeah that confuses me, because we are working with inequalities, there's no equality anywhere, so it seemed weird that they would state that in the solution. I guess they mentioned it like a special case. I understand now thanks for your answers.
          $endgroup$
          – Pero
          Dec 10 '18 at 23:19


















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