Chủ đề này có 2 trả lời

[frazier]

bài này dối xứng
$\sqrt[2]{\dfrac{a^3}{a^3 + (b + c)^3}}$ + $\sqrt[2]{\dfrac{b^3}{b^3 + (a + c)^3}}$ + $\sqrt[2]{\dfrac{c^3}{c^3 + (b + a)^3}}$ $\geq 1$

Bài viết đã được chỉnh sửa nội dung bởi frazier: 08-10-2009 - 21:50

[Janienguyen]

bài này dối xứng
$\sqrt[2]{\dfrac{a^3}{a^3 + (b + c)^3}}$ + $\sqrt[2]{\dfrac{b^3}{b^3 + (a + c)^3}}$ + $\sqrt[2]{\dfrac{c^3}{c^3 + (b + a)^3}}$ $\geq 1$

dug hold

[math93]

bài này đối xứng
$\sqrt{\dfrac{a^3}{a^3 + (b + c)^3}}+\sqrt{\dfrac{b^3}{b^3 + (a + c)^3}}+ \sqrt{\dfrac{c^3}{c^3 + (b + a)^3}}\geq 1$

Cách này khá hay:
Ta có:$\sqrt{1+x^3}\leq1+\dfrac{x^2}{2}, \forall x\geq 0$
Suy ra$\sqrt{\dfrac{a^3}{a^3 + (b + c)^3}}+\sqrt{\dfrac{b^3}{b^3 + (a + c)^3}}+ \sqrt{\dfrac{c^3}{c^3 + (b + a)^3}}$
$=\dfrac{1}{\sqrt{1+(\dfrac{b+c}{a})^3}}+\dfrac{1}{\sqrt{1+(\dfrac{a+c}{b})^3}}+\dfrac{1}{\sqrt{1+(\dfrac{b+a}{c})^3}}$
$\geq \dfrac{1}{1+\dfrac{1}{2}(\dfrac{b+c}{a})^2}+\dfrac{1}{1+\dfrac{1}{2}(\dfrac{a+c}{b})^2}+\dfrac{1}{1+\dfrac{1}{2}(\dfrac{b+a}{c})^2}$
$=\dfrac{a^2}{a^2+\dfrac{1}{2}(b+c)^2}+\dfrac{b^2}{b^2+\dfrac{1}{2}(a+c)^2}+\dfrac{c^2}{c^2+\dfrac{1}{2}(b+a)^2}$
$=\dfrac{a^2}{a^2+b^2+c^2}+\dfrac{b^2}{a^2+b^2+c^2}+\dfrac{c^2}{a^2+b^2+c^2}=1$