cho` :a;b;c>0` và `abc=1` `CM: 1/(\sqrt[ab+a+2])+1/(\sqrt[cb+b+2])+1/(\sqrt[ac+c+2]) ≤3/2`

Đáp án:

Giải thích các bước giải:

 Áp dụng CT với `x,y,z > 0` ta có: `x+y+z \le \sqrt{3(x^2+y^2+x^2)}`

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

– Với `x,y>0` ta có:

`x+y \ge 2\sqrt{xy}⇒ \frac{1}{x+y} \le \frac{1}{4}(\frac{1}{x}+\frac{1}{y})`

Áp dụng ta có:

`\frac{1}{ab+a+2}=\frac{1}{ab+1+a+1}=\frac{1}{ab+abc+a+1}=\frac{1}{ab(c+1)+(a+1)} \le \frac{1}{4}(\frac{1}{ab(c+1)}+\frac{1}{a+1})=\frac{1}{4}(\frac{abc}{ab(a+1)}+\frac{1}{a+1})=\frac{1}{4}(\frac{c}{c+1}+\frac{1}{a+1})`

`⇒ \frac{1}{ab+a+2} \le \frac{1}{4}(\frac{c}{c+1}+\frac{1}{a+1})`

Tương tự : `\frac{1}{bc+b+2} \le \frac{1}{4}(\frac{a}{a+1}+\frac{1}{b+1})`

`\frac{1}{ca+c+2} \le \frac{1}{4}(\frac{b}{b+1}+\frac{1}{c+1})`

`⇒ \sqrt{3(\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2})} \le \sqrt{3.\frac{1}{4}(\frac{c}{c+1}+\frac{1}{a+1}+\frac{a}{a+1}+\frac{1}{b+1}+\frac{b}{b+1}+\frac{1}{c+1})}=\frac{3}{2}`

Vậy `\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ca+c+2}} \le 3/2`

Dấu “=” xảy ra khi `a=b=c=1`