Question

cho a,b,c là các số thực dương
Chứng minh rằng: \(\dfrac{a+3c}{a+b}+\dfrac{a+3b}{a+c}+\dfrac{2\text{a}}{b+c}\ge5\)

Answer 1

Lời giải:

Ta có:

\(\text{VT}=\frac{a+c+2c}{a+b}+\frac{a+b+2b}{a+c}+\frac{2a}{b+c}\)

\(=\left(\frac{a+c}{a+b}+\frac{a+b}{a+c}\right)+2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)\)

Áp dụng BĐT AM-GM: \(\frac{a+c}{a+b}+\frac{a+b}{a+c}\geq 2\)

Và:

\(2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)=2\left(\frac{c+a+b}{a+b}+\frac{b+a+c}{a+c}+\frac{a+b+c}{b+c}-3\right)\)

\(=2(a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-6\)

\(=[(a+b)+(b+c)+(c+a)]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-6\)

\(\geq 3\sqrt[3]{(a+b)(b+c)(c+a)}.3\sqrt[3]{\frac{1}{(a+b)(b+c)(c+a)}}-6=9-6=3\)

Do đó:

\(\text{VT}\geq 2+3=5\)

Ta có đpcm

Dấu bằng xảy ra khi $a=b=c$