<s>[INST] Stronger version of AM-GM with condition If $a,b,c>0$ and $a^2+b^2+c^2=3$, prove that
$$a+b+c\ge 3(abc)^{13/76}.$$
I was thinking of using AM-GM :P but clearly some clever trick is needed. I was thinking about expanding but that is not feasible too. Note I would like a non-computational proof which can be done by hand. Also this was given to me by my teacher who has solved it and so I suppose it is correct. So any help?
 [/INST] First we check that $abc \leq 1$, it's easy by AM-GM:
$1=\frac{a^2+b^2+c^2}{3} \leq (abc)^{\frac{2}{3}}$
Next we prove that $(a+b+c)^2 \geq 9(abc)^{\frac{13*2}{76}}$. If we prove this the inequality $a+b+c \geq 3(abc)^{\frac{13}{76}}$ will be proven, because $x \leq y$ implies $\sqrt{x} \leq \sqrt{y}$ for $x,y \geq 0$.Next:
$(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=1+1+1+ab+ab+ac+ac+bc+bc$
Apply AM-GM to this sum. We have:
$ 1+1+1+ab+ab+ac+ac+bc+bc \geq 9(abc)^{\frac{2}{9}}$,
but:
$9(abc)^{\frac{2}{9}} \geq 9(abc)^{\frac{2*13}{76}}$, because it's equal:
$1 \geq (abc)^{\frac{2*13}{76}-\frac{2}{9}}=(abc)^{\frac{41}{342}}$ and it's true, because $abc \leq 1$, so
$(a+b+c)^2 \geq 9(abc)^{\frac{2*13}{76}}$
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