1.

Ta có:

\(x^4+y^4\ge\dfrac{1}{2}\left(x^2+y^2\right)^2=\dfrac{1}{2}\left(x^2+y^2\right)\left(x^2+y^2\right)\ge\left(x^2+y^2\right)xy\)

Đặt vế trái của BĐT cần chứng minh là P, áp dụng bồ đề vừa chứng minh ta có:

\(P\le\dfrac{a.abc}{bc\left(b^2+c^2\right)+a.abc}+\dfrac{b.abc}{ca\left(c^2+a^2\right)+b.abc}+\dfrac{c.abc}{ab\left(a^2+b^2\right)+c.abc}\)

\(P\le\dfrac{a^2.bc}{bc\left(a^2+b^2+c^2\right)}+\dfrac{b^2.ac}{ca\left(a^2+b^2+c^2\right)}+\dfrac{c^2.ab}{ab\left(a^2+b^2+c^2\right)}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

2.

\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}=1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)

\(\dfrac{X+1}{X-2}+\dfrac{X-2}{X+1}+\dfrac{X-14}{\left(X-2\right)\left(X+2\right)}\)

\(=\dfrac{\left(X+1\right)\left(X+1\right)\left(X+2\right)+\left(X-2\right)\left(X-2\right)\left(X+2\right)+\left(X-14\right)\left(X+1\right)}{\left(X-2\right)\left(X+2\right)\left(X+1\right)}\)

\(=\dfrac{X^3+2X^2+2X^2+4X+X+2+X^3+2X^2-4X^2-8X+4X+8+X^2+X-14X-14}{\left(X-2\right)\left(X+2\right)\left(X+1\right)}\)

\(=\dfrac{2X^3+3X^2-12X-4}{\left(X-2\right)\left(X+2\right)\left(X+1\right)}\)

\(\dfrac{x}{0}\)

Lời giải:
Ta có:

\(C=\frac{5(x^2-4x+4)-2x+5}{x^2-4x+4}=\frac{5(x-2)^2-2(x-2)+1}{(x-2)^2}=5-\frac{2}{x-2}+\frac{1}{(x-2)^2}\)

Đặt $\frac{1}{x-2}=t$ thì:

$C=t^2-2t+5=(t-1)^2+4\geq 4$ với mọi $t$

$\Rightarrow C_{\min}=4$. Vậy GTNN của $C$ là $4$. Giá trị này đạt tại $t=1$

$\Leftrightarrow \frac{1}{x-2}=1\Leftrightarrow x=3$