giúp

a)

Vì ${{a}_{1}}\le {{a}_{2}}$ nên ${{a}_{1}}-{{a}_{2}}\le 0$

Vì ${{b}_{1}}\le {{b}_{2}}$ nên ${{b}_{1}}-{{b}_{2}}\le 0$

$\to \left( {{a}_{1}}-{{a}_{2}} \right)\left( {{b}_{1}}-{{b}_{2}} \right)\ge 0$

$\to {{a}_{1}}{{b}_{1}}-{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}\ge 0$

$\to 2\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}} \right)\ge {{a}_{1}}{{b}_{1}}+{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}$

$\to 2\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}} \right)\ge {{a}_{1}}\left( {{b}_{1}}+{{b}_{2}} \right)+{{a}_{2}}\left( {{b}_{1}}+{{b}_{2}} \right)$

$\to 2\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}} \right)\ge \left( {{a}_{1}}+{{a}_{2}} \right)\left( {{b}_{1}}+{{b}_{2}} \right)$

$\to \dfrac{{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}}{2}\ge \dfrac{{{a}_{1}}+{{a}_{2}}}{2}\cdot \dfrac{{{b}_{1}}+{{b}_{2}}}{2}$

b)

Tương tự câu a, chỉ là đổi chiều

Vì ${{a}_{1}}\le {{a}_{2}}$ nên ${{a}_{1}}-{{a}_{2}}\le 0$

Vì ${{b}_{1}}\ge {{b}_{2}}$ nên ${{b}_{1}}-{{b}_{2}}\ge 0$

$\to \left( {{a}_{1}}-{{a}_{2}} \right)\left( {{b}_{1}}-{{b}_{2}} \right)\le $

$\to {{a}_{1}}{{b}_{1}}-{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}\le 0$

$\to 2\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}} \right)\le {{a}_{1}}{{b}_{1}}+{{a}_{1}}{{b}_{2}}+{{a}_{2}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}$

$\to 2\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}} \right)\le {{a}_{1}}\left( {{b}_{1}}+{{b}_{2}} \right)+{{a}_{2}}\left( {{b}_{1}}+{{b}_{2}} \right)$

$\to 2\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}} \right)\le \left( {{a}_{1}}+{{a}_{2}} \right)\left( {{b}_{1}}+{{b}_{2}} \right)$

$\to \dfrac{{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}}{2}\le \dfrac{{{a}_{1}}+{{a}_{2}}}{2}\cdot \dfrac{{{b}_{1}}+{{b}_{2}}}{2}$

c)

Vì $x,y\ge 0$

Giả sử $x\le y$

Thì khi đó ${{x}^{91}}\le {{y}^{91}}$  và  ${{x}^{9}}\le {{y}^{9}}$

Áp dụng câu a,

Ta có: $\dfrac{{{x}^{91}}.{{x}^{9}}+{{y}^{91}}.{{y}^{9}}}{2}\ge \dfrac{{{x}^{91}}+{{y}^{91}}}{2}\cdot \dfrac{{{x}^{9}}+{{y}^{9}}}{2}$

$\to \dfrac{{{x}^{100}}+{{y}^{100}}}{2}\ge \dfrac{{{x}^{91}}+{{y}^{91}}}{2}\cdot \dfrac{{{x}^{9}}+{{y}^{9}}}{2}$