# Nice inequality

• Let $a,b,c$ are positive real number,$abc \geq 1$.Prove that:
$\frac{a^3}{a^3+a^2+a+1}+\frac{b^3}{b^3+b^2+b+1}+\frac{c^3}{c^3+c^2+c+1}\geq \frac{3}{4}$

• Penser à C.S et la méthode pqr

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• This post is deleted!

• On utilisera les notations $p=a+b+c$ et $q=ab+ac+bc$ et $r=abc$
D apres $C.S$ $L.H.S\geq \frac{(a+b+c)^2}{p^2+p-2q +\frac{q}{r}+3}= \frac{p^2}{p^2+p-2q +\frac{q}{r}+3}$
Donc il suffit de montrer que $4p^2 \geq$ $3 (p^2+p-2q +\frac{q}{r}+3)$
Equivalent a $p^2+6q \geq 3p +3 \frac{q}{r} +9$
Or on a $p\geq 3(abc)^{1/3} =3$ donc $p^2\geq 3p$
et $3q \geq 3.3(abc)^{2/3} =9$
et $3q\geq \frac{3q}{r}$ Car $r\geq 1$
$CQFD$

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