Câu IV (1 điểm). Với a, b, c là những số thực dương. Chứng minh rằng$\frac{a}{4c}$ + $\frac{c}{4b}$ + $\frac{a+b}{2a+2b}$ + $\frac{3a+c}{a+3b}$ $\geq$ $\frac{16a}{5a+3b}$ 

Giải thích các bước giải:

Ta có:

$\dfrac{a}{4c}+\dfrac{c}{4b}+\dfrac{a+b}{2a+2c}+\dfrac{3a+c}{a+3b}$

$\ge 2\sqrt{\dfrac{a}{4c}\cdot\dfrac{c}{4b}}+\dfrac{a+b}{2(a+c)}+\dfrac{a+c}{a+3b}+\dfrac{2a}{a+3b}$ 

$\ge \dfrac{a}{2\sqrt{ab}}+\dfrac{a+b}{2(a+c)}+\dfrac{a+c}{a+3b}+\dfrac{2a}{a+3b}$ 

$\ge \dfrac{a}{a+b}+\dfrac{a+b}{2(a+c)}+\dfrac{a+c}{a+3b}+\dfrac{2a}{a+3b}$ 

$\ge 3\sqrt[3]{\dfrac{a}{a+b}\cdot\dfrac{a+b}{2(a+c)}\cdot\dfrac{a+c}{a+3b}}+\dfrac{2a}{a+3b}$ 

$\ge 3\sqrt[3]{\dfrac{a}{2(a+3b)}}+\dfrac{2a}{a+3b}$ 

$\ge \dfrac{3a}{\sqrt[3]{2a^2(a+3b)}}+\dfrac{2a}{a+3b}$ 

$\ge \dfrac{6a}{2\sqrt[3]{2a^2(a+3b)}}+\dfrac{2a}{a+3b}$ 

$\ge \dfrac{6a}{\sqrt[3]{2^3\cdot 2a^2(a+3b)}}+\dfrac{2a}{a+3b}$ 

$\ge \dfrac{6a}{\sqrt[3]{4a\cdot 4a\cdot (a+3b)}}+\dfrac{2a}{a+3b}$ 

$\ge \dfrac{18a}{3\sqrt[3]{4a\cdot 4a\cdot (a+3b)}}+\dfrac{2a}{a+3b}$ 

$\ge \dfrac{18a}{4a+4a+(a+3b)}+\dfrac{2a}{a+3b}$ 

$\ge \dfrac{18a}{9a+3b}+\dfrac{2a}{a+3b}$ 

$\ge 2a\cdot (\dfrac{9}{9a+3b}+\dfrac{1}{a+3b})$ 

$\ge 2a\cdot (\dfrac{3^2}{9a+3b}+\dfrac{1}{a+3b})$ 

$\ge 2a\cdot \dfrac{(3+1)^2}{(9a+3b)+(a+3b)}$ 

$\ge 2a\cdot \dfrac{16}{10a+6b}$

$\ge \dfrac{16}{5a+3b}$

$\to đpcm$