a) Xét hiệu: $ \dfrac{1}{x + y} - \dfrac{1}{4}\bigg(\dfrac{1}{x} + \dfrac{1}{y}\bigg)(x,y > 0)$
$=\dfrac{1}{x+y} - \dfrac{1}{4x} - \dfrac{1}{4y}$
$=\dfrac{4xy - y(x + y) - x(x+y)}{4xy(x+y)}$
$=\dfrac{4xy - xy - y^2 - x^2 - xy}{4xy(x+y)}$
$=\dfrac{-(x^2 - 2xy + y^2)}{4xy(x+y)}$
$=\dfrac{-(x - y)^2}{4xy(x+y)}$
Ta có: $-(x-y)^2 \leq 0 ∀ x,y$
Mà $4xy(x+y) > 0$ (Vì x,y > 0)$
Do đó: $\dfrac{-(x - y)^2}{4xy(x+y)} \leq 0 ∀ x,y >0$
hay $\dfrac{1}{x + y} - \dfrac{1}{4}\bigg(\dfrac{1}{x} + \dfrac{1}{y}\bigg) \leq 0$
⇔ $\dfrac{1}{x + y} \leq \dfrac{1}{4}\bigg(\dfrac{1}{x} + \dfrac{1}{y}\bigg)$
b) Đề hơi sai nên mình sửa lại ạ
Ta có: $A = \dfrac{ab}{c+1} + \dfrac{bc}{a+1} + \dfrac{ac}{b+1}$
⇒ $A = \dfrac{ab}{c + a + b + c} + \dfrac{bc}{a + b + c + a} + \dfrac{ac}{b + a + b + c}$ (Vì a + b + c = 1)$
Áp dụng bất đẳng thức ở ý a) cho hai số dương ta có:
$\dfrac{1}{c + a + b+ c} \leq \dfrac{1}{4}\bigg(\dfrac{1}{c + a} + \dfrac{1}{b + c}\bigg)$
⇒ $\dfrac{ab}{c + a + b+ c} \leq \dfrac{ab}{4\bigg(\dfrac{1}{c +a} + \dfrac{1}{b+c}\bigg)}$
Tương tự ta có : $\dfrac{bc}{a + b + c + a} \leq \dfrac{bc}{4\bigg(\dfrac{1}{a + b} + \dfrac{1}{c +a}\bigg)}$
$\dfrac{ac}{b +a + b + c } \leq \dfrac{ac}{4\bigg(\dfrac{1}{a + b} + \dfrac{1}{b+c}\bigg)}$
Cộng vế với vế ta được:
$\dfrac{1}{c + a + b+ c}+\dfrac{bc}{a + b + c + a}+ \dfrac{ac}{b +a + b + c } \leq \dfrac{1}{4}\bigg(\dfrac{ab}{c + a} + \dfrac{ab}{b+c} + \dfrac{bc}{a+b} + \dfrac{bc}{a+c} + \dfrac{ac}{a+b} + \dfrac{ac}{b+c}\bigg)$
hay $A \leq \dfrac{1}{4}(a + b +c)$
⇒ $A \leq \dfrac{1}{4}$ (Vì a + b + c = 1)
