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If $a$, $b$ and $c$ are three positive real numbers, prove that $$\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$$

I think we are supposed to use the AM-GM inequality here; or maybe this could even be solved using the $T_2$'s Lemma, although I'm a bit skeptical about that.

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  • $\begingroup$ Or you can use AM-GM on the numerators to get Nesbitt. $\endgroup$ – Macavity Oct 6 '16 at 1:35
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Let us define $t=a+b+c$. First we note that

$$\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b}= \left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+ \left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)$$

so by Titu's lemma:

$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq\frac{(a+b+c)^2}{2(a+b+c)}=\frac{t}{2}$$

and

$$\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\geq\frac{3^2}{2(a+b+c)}=\frac{9}{2t}.$$

Therefore, we have to check that

$$\frac{t}{2}+\frac{9}{2t}\geq3$$

for $t>0$. But this follows from the fact that the polynomial $p(t)=t^2-6t+9=(t-3)^2\geq0$.

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After using AM-GM we need to prove that $$(a^2+1)(b^2+1)(c^2+1)\geq(a+b)(a+c)(b+c)$$ which follows from C-S: $(a^2+1)(1+b^2)\geq(a+b)^2$.

Done!

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