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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.

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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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