# Equality in the A.M G.M inequality

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 $$a_1=a_2=a_3\cdots =a_n$$.

• Does this answer your question? A proof for AM-GM inequality (Apostol: Mathematical Analysis) May 18 at 20:11
• 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. May 18 at 20:14
• @callculus42 he is saying that the AM=GM not the inverse. May 18 at 20:15
• @Youem Oh yes. Thanks. May 18 at 20:18

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.

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?

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