Question

Xét a,b,c thỏa mãn a + b + c = 6. Chứng minh rằng \(\dfrac{a}{3a+1}+\dfrac{b}{3b+1}+\dfrac{c}{3c+1}\)≤ \(\dfrac{6}{7}\)

Answer 1

Lời giải:

$3\text{VT}=\frac{3a}{3a+1}+\frac{3b}{3b+1}+\frac{3c}{3c+1}$

$=1-\frac{1}{3a+1}+1-\frac{1}{3b+1}+1-\frac{1}{3c+1}$

$=3-\left[\frac{1}{3a+1}+\frac{1}{3b+1}+\frac{1}{3c+1}\right]$
Áp dụng BĐT Cauchy-Schwarz:

$\frac{1}{3a+1}+\frac{1}{3b+1}+\frac{1}{3c+1}\geq \frac{9}{3a+1+3b+1+3c+1}=\frac{9}{3(a+b+c)+3}=\frac{9}{3.6+3}=\frac{3}{7}$

$\Rightarrow 3\text{VT}\leq 3-\frac{3}{7}=\frac{18}{7}$

$\Rightarrow \text{VT}\leq \frac{6}{7}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$