Question

[nguyễn thị kim oanh]

Please, help me!!!!!!!!!!!!!!!

Answer 1

[Nguyễn Thái Thịnh]

Dễ dàng chứng minh BĐT \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\forall a,b,c,d\)

Hay \(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\ge\sqrt{\left(x+y\right)^2+\left(\frac{1}{x}+\frac{1}{y}\right)^2}\)

\(+\sqrt{z^2+\frac{1}{z^2}}\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)

Đặt \(\left(x+y+z\right)^2=t\Leftrightarrow0< t\le\frac{9}{4}\)

Vì \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\ge\frac{81}{\left(x+y+z\right)^2}\) nên \(\left(x+y+z\right)^2\ge t+\frac{81}{t}\)

Mà hàm số \(f\left(t\right)=t+\frac{81}{t}< 0\) trên khoảng \(0< t\le\frac{9}{4}\) nên \(f\left(t\right)_{min}\) là \(f\left(\frac{9}{4}\right)=\frac{9.17}{4}=\frac{153}{4}\)

Do đó \(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\ge f\left(t\right)\ge\frac{3}{2}\sqrt{17}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{2}\)