Áp dụng BĐT Cô-si :

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\left|\frac{a}{c}\right|\ge\frac{a}{c}\)

\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\left|\frac{b}{a}\right|\ge\frac{b}{a}\)

\(\frac{c^2}{a^2}+\frac{a^2}{b^2}\ge2\left|\frac{c}{b}\right|\ge\frac{c}{b}\)

Cộng 3 vế của 3 đẳng thức trên với nhau có :

\(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\right)\)

\(\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\)

Vậy ...

Ta có \(\left(x-y\right)^2\ge0\Leftrightarrow x^2-2xy+y^2\)

                                           \(\Leftrightarrow x^2+y^2\ge0\)

Áp dụng bài toán trên, ta có

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\frac{ab}{bc}\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\frac{a}{c}\)               (1)

Chứng minh tương tự, ta được

\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\frac{b}{a}\)                   (2)

\(\frac{c^2}{a^2}+\frac{a^2}{b^2}\ge2\frac{c}{b}\)                    (3)

Cộng (1)(2)(3), ta được

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}\ge2\frac{a}{c}+2\frac{b}{a}+2\frac{c}{b}\)

\(\Leftrightarrow2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\)

\(\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)             \(\left(đpcm\right)\)

OIMEOI

may lam dai the thi cho no giai ho a 

Ta có: x² - 2xy - 4z² + y²
= (x² - 2xy + y²) - (2z)²
= (x - y)² - (2z)² 

=(x - y - 2z)(x- y +2z)    (*)
Thay x= 6; y= -4; z=45 vào biểu thức (*), ta đc:
(6 + 4 - 2.45)(6 + 4 +2.45)
= -80.100
=-8000
Vậy...

giúp mink di mink cho 2 l*i*k*e

To you :)

Ta có :

\(A=-x^2+7x\)

\(=-x^2+2.x.\frac{7}{2}-\frac{49}{4}+\frac{49}{4}\)

\(=\frac{49}{4}-\left(x^2-2.x.\frac{7}{2}+\frac{49}{4}\right)\)

\(=\frac{49}{4}-\left(x-\frac{7}{2}\right)^2\)

\(\left(x-\frac{7}{2}\right)^2\ge0\)

\(\Rightarrow\frac{49}{4}-\left(x-\frac{7}{2}\right)^2\le\frac{49}{4}\)

\(\Rightarrow Min_A=\frac{49}{4}\Leftrightarrow x=\frac{7}{2}\)

Vậy ...