Question

 Cho 3 số dương x,y,z là 3 số dương và x+y+z=1. CMR: \(\sqrt{x^2+\frac{1}{x^2}}\)+\(\sqrt{y^2+\frac{1}{y^2}}\)+\(\sqrt{z^2+\frac{1}{z^2}}\)\(\ge\sqrt{82}\)

Answer 1

Gọi \(\overrightarrow{1a}=\left(x;\frac{1}{x}\right);\overrightarrow{b}=\left(y;\frac{1}{y}\right);\overrightarrow{c}=\left(z;\frac{1}{z}\right)\)

Ta có:

\(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}=\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|\)

\(\ge\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|=\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)\(\ge\sqrt{1^2+\frac{9^2}{\left(x+y+z\right)^2}}\)

\(=\sqrt{1+81}=\sqrt{82}\)

    

Answer 2

Áp dụng BDT MInkopki

VT\(\ge\)\(\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)\(\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}=\sqrt{82}\)

BDT minkopki

\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}+\sqrt{e^2+f^2}\ge\sqrt{\left(a+c+e\right)^2+\left(b+d+f\right)^2}\)

Answer 3

Áp dụng bất đẳng thức bunhiacopxki ta có

\(\sqrt{\left(x^2+\frac{1}{x^2}\right)\left(1+81\right)}\ge x+\frac{9}{x}\)

Tương tự với các trường hợp còn lại

Đặt VT =A

Ta có \(\sqrt{82}.A\ge x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\ge1+\frac{9^2}{x+y+z}=82\)

=> \(A\ge\sqrt{82}\)

Answer 4

=√1+81=√82

100% đó.