<s>[INST] Solve inequality using AM-GM / Cauchy-Schwarz Given positive real numbers $a$, $b$, $c$ satisfying $a^2+b^2+c^2=3$. Prove that
$$\dfrac{1}{a^2+7}+\dfrac{1}{b^2+7}+\dfrac{1}{c^2+7} \leq \dfrac{1}{4} \left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right).$$
I have tried using $a+b\leq \sqrt{2\left(a^2+b^2\right)}=\sqrt{2\left(3-c^2\right)}$ but it goes to a wrong inequality. I know this problem can be solved by using AM-GM inequality (Cauchy). Please help me with this, thanks.
 [/INST] Let $a+b+c=3u.$
Thus, $u\leq1$ and by C-S $$\sum_{cyc}\frac{1}{a^2+7}=\sum_{cyc}\frac{b^2+c^2+6}{(a^2+1+1+5)(1+b^2+c^2+5)}\leq$$
$$\leq\sum_{cyc}\frac{b^2+c^2+6}{(a+b+c+5)^2}=\frac{24}{(a+b+c+5)^2}=\frac{24}{(3u+5)^2}.$$
In another hand, by C-S again:
$$\sum_{cyc}\frac{1}{a+b}=\frac{1}{2(a+b+c)}\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{a+b}\geq\frac{9}{2(a+b+c)}=\frac{3}{2u}.$$
Id est, it's enough to prove that
$$\frac{3}{8u}\geq\frac{24}{(3u+5)^2}$$ or
$$(1-u)(25-9u)\geq0,$$ which is obvious.
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