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Let $a_1, a_2,\cdots a_n$ be positive real numbers. The A.M-G.M inequality states that $$(a_1a_2\cdots a_n)^\frac{1}{n}\leq\frac{a_1+a_2+\cdots +a_n}{n}$$ with equality if and only if $a_1=a_2=a_3\cdots =a_n$.

I was able to prove the inequality using Lagrange Multipliers method. Can anyone help me prove the Equality part. i.e if A.M=G.M then $a_1=a_2=a_3\cdots =a_n$.

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    $\begingroup$ Does this answer your question? A proof for AM-GM inequality (Apostol: Mathematical Analysis) $\endgroup$
    – Kroki
    May 18 at 20:11
  • $\begingroup$ If they are equal then you have $\left(a_1^n\right)^{\frac1{n}}$ on the LHS and $\frac{n\cdot a_1}{n}$ on the RHS. $\endgroup$
    – callculus42
    May 18 at 20:14
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    $\begingroup$ @callculus42 he is saying that the AM=GM not the inverse. $\endgroup$
    – Kroki
    May 18 at 20:15
  • $\begingroup$ @Youem Oh yes. Thanks. $\endgroup$
    – callculus42
    May 18 at 20:18

2 Answers 2

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Assume that you have proved LHS $\le$ RHS by some way.

Now we prove that if LHS $=$ RHS, then $a_1 = a_2 = \cdots = a_n$.

Assume, for the sake of contradiction, that LHS = RHS for some positive real numbers $a_1, a_2, \cdots, a_n$, not all equal.

WLOG, assume that $a_1 < a_2$. Let $b_1 = b_2 = (a_1 + a_2)/2$. We have $b_1 + b_2 = a_1 + a_2$ and $$b_1b_2 - a_1a_2 = (a_1 - a_2)^2/4 > 0.$$ If we replace $a_1, a_2$ with $b_1, b_2$, we get LHS $>$ RHS. A contradiction.

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Hint: if you were able to do it by Lagrange Multipliers then AM$=$GM implies $\left(a_1,\ldots,a_n\right)$ satisfies the KKT. Can you continue from here?

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  • $\begingroup$ Can you please what you mean by KKT? $\endgroup$
    – user31459
    May 18 at 20:22
  • $\begingroup$ How did you manage to prove the inequality by Lagrange multipliers? $\endgroup$
    – Kroki
    May 18 at 20:23
  • $\begingroup$ ooOOOooo I need to go backwards and use that if the equality holds then the extremum is achieved $\endgroup$
    – user31459
    May 18 at 20:30
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    $\begingroup$ Yes! That's all. $\endgroup$
    – Kroki
    May 18 at 20:30

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