ôn vào lớp 10 BĐT part 1

Question

Cho

$a, b > 0$ và

$a^{9}+b^{9}=2$ . C/m :

$\frac{a^{2}}{b}+\frac{b^{2}}{a} \geq 2$

tran85295 · Answer 1 · 4/26/2016 10:50:09 PM

Cần cm :

$\frac{a^2}{b}+\frac {b^2}a \ge\sqrt[9]{(a^9+b^9).2^8}$

$\Leftrightarrow \frac{(a^3+b^3)^9}{(ab)^9} \ge2^8(a^9+b^9)$

$\Leftrightarrow \frac{(x+y)^9}{x^3y^3} \ge 2^8(x^3+y^3) \qquad(x=a^3,y=b^3)$

Đặt

$m=\frac{2x}{x+y},n=\frac{2y}{x+y}\Rightarrow x=\frac{m(x+y)}{2},y=\frac{n(x+y)}{2},m+n=2$

Bđt trên

$\Leftrightarrow \frac{(m+n)^9}{m^3n^3} \ge 2^8(m^3+n^3)$

$\Leftrightarrow \frac{2^9}{m^3n^3} \ge 2^8(m+n)(m^2-mn+n^2)$

$\Leftrightarrow m^3n^3[(m+n)^2-3mn] \le1$

$\Leftrightarrow m^3n^3(3mn-4) +1\ge 0\Leftrightarrow (t-1)^2(3t^2+2t+1) \ge0$ (luôn đúng )

$\qquad(t=mn)$

Vậy ta có đpcm @@

Sửa 26-04-16 10:50 PM

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Trả lời 26-04-16 10:47 PM

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