let $a,b,c>0$ show that $$\left(1+\dfrac{4a}{b+c}\right)\left(1+\dfrac{4b}{c+a}\right)\left(1+\dfrac{4c}{a+b}\right)\ge 25$$
It seem hard to prove AM-GM.Cauchy-Schwarz
How prove $\left(1+\frac{4a}{b+c}\right)\left(1+\frac{4b}{c+a}\right)\left(1+\frac{4c}{a+b}\right)\ge 25$
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$\begingroup$
inequality
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0Are $a,b,c$ real numbers or integers? – 2017-01-16
2 Answers
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Hint
On full expansion, this reduces to a combination of Schur's inequality $\sum a^3 +3abc\geqslant \sum ab(a+b)$ and $abc>0$, so the inequality is in fact strict.
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$$\Leftrightarrow a^{3}+b^3+c^3+7abc\geq ab(a+b)+bc(b+c)+ca(c+a) > 0$$
Right by Schur: $$a^{3}+b^3+c^3+7abc> a^{3}+b^3+c^3+3abc\geq ab(a+b)+bc(b+c)+ca(c+a)$$