nhờ chuyên gia,
CMR:
`a) a^3/{a^2+b^2}+b^3/{b^2+c^2}+c^3/{c^2+d^2}+d^3/{d^2+a^2}\ge{a+b+c+d}/2` với a,b,c,d dương
`b) a^4/{a^3+2b^3}+b^4/{b^3+2c^3}+c^4/{c^3+2d^3}+d^4/{d^3+2a^2}\ge{a+b+c+d}/3` với a,b,c,d dương
nhờ chuyên gia,
CMR:
`a) a^3/{a^2+b^2}+b^3/{b^2+c^2}+c^3/{c^2+d^2}+d^3/{d^2+a^2}\ge{a+b+c+d}/2` với a,b,c,d dương
`b) a^4/{a^3+2b^3}+b^4/{b^3+2c^3}+c^4/{c^3+2d^3}+d^4/{d^3+2a^2}\ge{a+b+c+d}/3` với a,b,c,d dương
a) Ta có:
$\dfrac{a^3}{a^2 + b^2} = \dfrac{a^3 + ab^2 – ab^2}{a^2 + b^2} = \dfrac{a(a^2 + b^2) – ab^2}{a^2 + b^2} = a – \dfrac{ab^2}{a^2 + b^2}$
Ta lại có:
$a^2 + b^2 \geq 2\sqrt{a^2b^2}$
$\Rightarrow \dfrac{ab^2}{a^2 + b^2} \leq \dfrac{ab^2}{2\sqrt{a^2b^2}}$
$\Rightarrow a – \dfrac{ab^2}{a^2 + b^2} \geq a – \dfrac{ab^2}{2\sqrt{a^2b^2}} = a – \dfrac{b}{2}$
$\Rightarrow \dfrac{a^3}{a^2 + b^2} \geq a – \dfrac{b}{2}$
Chứng minh tương tự, ta được:
$\dfrac{b^3}{b^2 + c^2} \geq b – \dfrac{c}{2}$
$\dfrac{c^3}{c^2 + d^2} \geq c – \dfrac{d}{2}$
$\dfrac{d^3}{d^2 + a^2} \geq d – \dfrac{a}{2}$
Cộng vế theo vế, ta được:
$\dfrac{a^3}{a^2 + b^2} + \dfrac{b^3}{b^2 + c^2} + \dfrac{c^3}{c^2 + d^2} + \dfrac{d^3}{d^2 + a^2} \geq a – \dfrac{b}{2} + b – \dfrac{c}{2}+c – \dfrac{d}{2}+d – \dfrac{a}{2} = \dfrac{a +b + c +d}{2}$
Dấu = xảy ra khi $a =b = c = d$
2) Ta có:
$\dfrac{a^4}{a^3 + 2b^3} = \dfrac{a^4 + 2ab^3 – 2ab^3}{a^3 + 2b^3} = \dfrac{a(a^3 + 2b^3)}{a^3 + 2b^3} – \dfrac{2ab^3}{a^3 + 2b^3} = a – \dfrac{2ab^3}{a^3 + 2b^3}$
Ta lại có:
$a^3 +2b^3 = a^3 + b^3 + b^3 \geq 3\sqrt[3]{a^3b^3b^3} = 3ab^2$
$\Rightarrow \dfrac{2ab^3}{a^3 + 2b^3} \leq \dfrac{2ab^3}{3ab^2} = \dfrac{2b}{3}$
$\Rightarrow a – \dfrac{2ab^3}{a^3 + 2b^3} \geq a – \dfrac{2b}{3}$
$\Rightarrow \dfrac{a^4}{a^3 + 2b^3} \geq a – \dfrac{2b}{3}$
Chứng minh tương tự, ta được:
$\dfrac{b^4}{b^3 + 2c^3} \geq b – \dfrac{2c}{3}$
$\dfrac{c^4}{c^3 + 2d^3} \geq c – \dfrac{2d}{3}$
$\dfrac{d^4}{d^3 + 2a^3} \geq d – \dfrac{2a}{3}$
Cộng vế theo vế, ta đươc:
$ \dfrac{a^4}{a^3 + 2b^3} + \dfrac{b^4}{b^3 + 2c^3} + \dfrac{c^4}{c^3 + 2d^3} + \dfrac{d^4}{d^3 + 2a^3} \geq a – \dfrac{2b}{3} + b – \dfrac{2c}{3} + c – \dfrac{2d}{3} + d – \dfrac{2a}{3} = \dfrac{a + b + c + d}{3}$
Dấu = xảy ra khi $a = b = c =d$