Question

[ivyuyen]

 cho a , b , c > 0 thỏa mãn abc = 1 . chứng minh\(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ca+c+2}}\)  \(\le\frac{3}{2}\)

Answer 1

[Kiệt Nguyễn]

Vì abc = 1 nên ta có thể đặt \(\left(a;b;c\right)\rightarrow\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)\). Khi đó: 

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

\(\Rightarrow VT^2\le\left(1+1+1\right)\left(\Sigma_{cyc}\frac{yz}{xy+xz+2yz}\right)\left(\text{ }\right)\)(Theo BĐT Cauchy-Schwarz)

\(\le\frac{3}{4}\left[\Sigma_{cyc}yz\left(\frac{1}{xy+yz}+\frac{1}{xz+yz}\right)\right]=\frac{3}{4}\left(\Sigma_{cyc}\frac{xy+yz}{xy+yz}\right)=\frac{9}{4}\)

\(\Rightarrow VT\le\frac{3}{2}\)

Đẳng thức xảy ra khi x = y = z hay a = b = c = 1