Cho các số thực a, b, c thỏa mãn abc = 1, Chứng minh rằng: ab/a^4 + b^4 + ab

Ta có: $a^4+b^4\ge ab\left(a^2+b^2\right)$$\Rightarrow \frac{ab}{a^4+b^4+ab}\le \frac{ab}{ab\left(a^2+b^2\right)+ab}=\frac{1}{a^2+b^2+1}$

Tương tự có: $\frac{bc}{b^4+c^4+bc}\le \frac{1}{b^2+c^2+1}$;$\frac{ca}{c^4+a^4+ca}\le \frac{1}{c^2+a^2+1}$

Suy ra $VT\le \frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}$

Đặt $a^2=x^3;b^2=y^3\text{'}{c}^2=z^3$ ta có: xyz = 1 ( do abc = 1)

Suy ra: $VT\le \frac{1}{x^3+y^3+1}+\frac{1}{y^3+z^3+1}+\frac{1}{z^3+x^3+1}$

Dễ cm đc $x^3+y^3\ge xy\left(x+y\right)$

$VT\le \frac{1}{xy\left(x+y\right)+1}+\frac{1}{yz\left(y+z\right)+1}+\frac{1}{zx\left(z+x\right)+1}$

$VT\le \frac{z}{xyz\left(x+y\right)+z}+\frac{x}{xyz\left(y+z\right)+x}+\frac{y}{zxy\left(z+x\right)+y}$

$VT\le \frac{z}{x+y+z}+\frac{x}{x+y+z}+\frac{y}{zx+y+z}=1$

Vậy $VT\le 1$ Dấu “=” xảy ra khi a = b = c.