Question

[Khởi My]

Cho a,b,c >0 thỏa mãn abc=1. Chứng minh:

\(\dfrac{1}{\sqrt{ab+a+2}}+\dfrac{1}{\sqrt{bc+b+2}}+\dfrac{1}{\sqrt{ca+c+2}}\le\dfrac{3}{2}\)

Answer 1

[Akai Haruma]

Lời giải:
Vì $abc=1$ nên tồn tại $x,y,z$ sao cho : \((a,b,c)=\left(\frac{x}{y}, \frac{y}{z}, \frac{z}{x}\right)\)

Khi đó:

\(\text{VT}=\frac{1}{\sqrt{\frac{x}{z}+\frac{x}{y}+2}}+\frac{1}{\sqrt{\frac{y}{x}+\frac{y}{z}+2}}+\frac{1}{\sqrt{\frac{z}{y}+\frac{z}{x}+2}}=\frac{\sqrt{yz}}{\sqrt{xy+xz+2yz}}+\frac{\sqrt{xz}}{\sqrt{xy+yz+2xz}}+\frac{\sqrt{xy}}{\sqrt{xz+yz+2xy}}\)

Áp dụng BĐT Cauchy-Schwarz:

\(\text{VT}^2\leq (1+1+1)\left(\frac{yz}{xy+xz+2yz}+\frac{xz}{xy+yz+2xz}+\frac{xy}{xz+yz+2xy}\right)\)

\(\leq 3\left[\frac{yz}{4}\left(\frac{1}{xy+yz}+\frac{1}{xz+yz}\right)+\frac{xz}{4}\left(\frac{1}{xy+xz}+\frac{1}{xz+yz}\right)+\frac{xy}{4}\left(\frac{1}{xz+xy}+\frac{1}{yz+xy}\right)\right]\)

hay \(\text{VT}^2\leq \frac{3}{4}.\left(\frac{xy+yz}{xy+yz}+\frac{xy+xz}{xy+xz}+\frac{yz+xz}{yz+xz}\right)=\frac{9}{4}\)

\(\Rightarrow \text{VT}\leq \frac{3}{2}\) (đpcm)

Dấu "=" xảy ra khi $x=y=z$ hay $a=b=c=1$