Cho a,b,c dương.Chứng minh rằng
a.$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \sqrt{\frac{a^2+b^2}{2}}+\sqrt{\frac{b^2+c^2}{2}}+\sqrt{\frac{a^2+c^2}{2}} $
b.$(a^2+2)(b^2+2)(c^2+2) \geq 9(ab+bc+ca)$
c.$\left ( 1+\frac{a}{b} \right )\left ( 1+\frac{b}{c} \right )\left ( 1+\frac{c}{a} \right )\geq 2\left ( 1+\frac{a+b+c}{\sqrt[3]{abc}} \right ) $
a. Áp dụng Cauchy-Schwarz ta có :
$\sum \sqrt{\frac{a^2+b^2}{2}}\leq \sqrt{3\sum a^2}\leq \sqrt[4]{27(a^4+b^4+c^4)}$
Nên ta chỉ cần chứng minh
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \sqrt[4]{27(a^4+b^4+c^4)}$
Áp dụng BĐT Cauchy-Schwarz ta có :
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \frac{(a^2+b^2+c^2)^2}{a^2b+b^2c+c^2a}$
Lại có :
$a^2b+b^2c+c^2a\leq \sqrt{(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)}$
$\Rightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \frac{(a^2+b^2+c^2)^\frac{3}{2}}{\sqrt{(a^2b^2+b^2c^2+c^2a^2)}}$
Áp dụng AM-GM ta có
$(a^2+b^2+c^2)^2=(a^4+b^4+c^4)+(a^2b^2+b^2c^2+c^2a^2)+(a^2b^2+b^2c^2+c^2a^2)\geq 3\sqrt[3]{(a^4+b^4+c^4)(a^2b^2+b^2c^2+c^2a^2)^2}$
$\Rightarrow (a^2+b^2+c^2)^\frac{3}{2}\geq \sqrt{(a^2b^2+b^2c^2+c^2a^2).\sqrt{27(a^4+b^4+c^4)}}$
$\Rightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \sqrt[4]{27(a^4+b^4+c^4)}$