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$$\Leftrightarrow m^3n^3(3mn-4) +1\ge 0\Leftrightarrow (t-1)^2(3t^2+2t+1) \ge0$ (luôn đúng ) $\qquad(t=mn)$Vậy ta có đpcm @@
Cần cm : $\frac{a^2}{b}+\frac {b^2}a \ge\sqrt[9]{(a^9+b^9).2^8}$$\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)$$\Leftrightarrow m^3n^3[(m+n)^3-3mn] \le1$$\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 @@
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 @@