Bài 100703

Question

Chứng minh rằng:
1/

$\left ( ax+by \right )^{2}\leq \left ( a^{2}+ b^{2}\right ) \left ( x^{2}+ y^{2}\right ), \forall x,y,a,b\in R$
2/

$\left ( \frac{12}{5} \right )^{x}+ \left ( \frac{15}{4} \right )^{x}+ \left ( \frac{20}{3} \right )^{x} \geq 3^{x}+ 4^{x} + 5^{x} ,\forall x\in R$

score 0 · Answer 1

1/

$\left ( ax+by \right )^{2}\leq \left ( a^{2}+ b^{2} \right ) \left ( x^{2}+ y^{2} \right )$

$\Leftrightarrow \left ( ax \right )^{2}+2abxy+ \left ( by \right )^{2} \leq \left ( ax \right )^{2} + \left ( ay \right )^{2} + \left ( bx \right )^{2}+ \left ( by \right )^{2}$

$\Leftrightarrow \left ( ay \right )^{2} + \left ( bx \right )^{2} -2abxy\geq 0$

$\Leftrightarrow \left ( ay-bx \right )^{2} \geq 0$ ,luôn đúng

$\Rightarrow$ (ĐPCM)
2/Đặt:

$\begin{cases}\alpha=\left ( \frac{12}{5} \right )^{\frac{x}{2}}>0 \\ \beta=\left ( \frac{15}{4} \right )^{\frac{x}{2}}>0 \\ z=\left ( \frac{20}{3} \right )^{\frac{x}{2}}>0 \end{cases}$
Bất đẳng thức đã cho