hthang0030

$a^{3}+b^{3}+c^{3}\geq ab(a+b)+c^{3}\geq \sqrt{(a+b).abc.c^{2}}=2c\sqrt{\frac{9}{4}(a+b)}=3c\sqrt{a+b}$

Tương tự:
$a^{3}+b^{3}+c^{3}\geq 3b\sqrt{a+c}$

$a^{3}+b^{3}+c^{3}\geq 3a\sqrt{b+c}$

= > $a^{3}+b^{3}+c^{3}\geq c\sqrt{a+b} + b\sqrt{a+c} + a\sqrt{b+c}$

Dấu bằng xảy ra <=>a=b=c(Vô lí) Vậy $a^{3}+b^{3}+c^{3}> c\sqrt{a+b} + b\sqrt{a+c} + a\sqrt{b+c}$

$P=\frac{\sqrt{ab}}{c+ab}+\frac{\sqrt{bc}}{a+bc}+\frac{\sqrt{ac}}{b+ac} <=>P=\frac{\sqrt{ab}}{c(a+b+c)+ab}+\frac{\sqrt{bc}}{a(a+b+c)+bc}+\frac{\sqrt{ac}}{b(a+b+c)+ac}<=>P=\frac{\sqrt{ab}}{(c+a)(c+b)}+\frac{\sqrt{bc}}{(a+b)(a+c)}+\frac{\sqrt{ac}}{(b+a)(b+c)}<=>P=\frac{(a+b)\sqrt{ab}+(b+c)\sqrt{bc}+(a+c)\sqrt{ac}}{(a+b)(b+c)(c+a)}\geq \frac{2ab+2bc+2ac}{(1-a)(1-b)(1-c)}=\frac{2ab+2bc+2ac}{ab+bc+ac-abc}.Vì  \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}= 9 =>ab+ac+bc\geq 9abc=>\frac{2ab+2bc+2ac}{ab+bc+ac-abc}\geq \frac{9}{4}$

Vậy Min P=$\frac{9}{4}$ tại a=b=c=$\frac{1}{3}$

Cho 3 số không âm a,b,c thỏa mãn a²+b²+c²=3 và $a\geq b\geq c$.Chứng minh rằng:

$\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2}\leq 1$