+, Nếu \(x<1\) thì:

\(x^8-x^7+x^2-x+1\)

\(=x^8+x^2\left(1-x^5\right)+\left(1-x\right)\)

\(>0\)

+, Nếu \(x\ge1\) thì:

\(x^8-x^7+x^2-x+1\)

\(=x^7\left(x-1\right)+x\left(x-1\right)+1\)

\(>0\)

+, Do đó \(x^8-x^7+x^2-x+1>0\forall x\) ( đpcm )

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

\(=\frac12\left(\frac{2a^2}{b^2}+\frac{2b^2}{c^2}+\frac{2c^2}{a^2}\right)\)

\(=\frac12\left\lbrack\left(\frac{a^2}{b^2}-\frac{2a}{c}+\frac{b^2}{c^2}\right)+\left(\frac{b^2}{c^2}-\frac{2b}{a}+\frac{c^2}{a^2}\right)+\left(\frac{c^2}{a^2}-\frac{2c}{b}+\frac{a^2}{b^2}\right)+2\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\right\rbrack\)

\(=\frac12\left\lbrack\left(\frac{a}{b}-\frac{b}{c}\right)^2+\left(\frac{b}{c}-\frac{c}{a}\right)^2+\left(\frac{c}{a}-\frac{a}{b}\right)^2\right\rbrack+\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\)

\(\ge\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\) ( đpcm )

( Dấu "=" xảy ra khi \(a=b=c\) )

+, Do \(\begin{cases}x>\sqrt2\\ y>\sqrt2\end{cases}\) nên \(\begin{cases}x^2>2\\ y^2>2\end{cases}\) thì \(\begin{cases}x^5>2x^3\\ y^5>2y^3\end{cases}\) hay \(x^5+y^5>2\left(x^3+y^3\right)\)

+, \(x^4-x^3y+x^2y^2-xy^3+y^4\)

\(=\frac{\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)}{x+y}\)

\(=\frac{x^5+y^5}{x+y}\)

\(>\frac{2\left(x^3+y^3\right)}{x+y}\)

\(=\frac{2\left(x+y\right)\left(x^2-xy+y^2\right)}{x+y}\)

\(=\frac{2\left(x+y\right)\left(x^2+y^2\right)-2xy\left(x+y\right)}{x+y}\)

\(=\frac{\left(x+y\right)\left(x^2+y^2\right)+\left(x+y\right)\left(x^2+y^2\right)-2xy\left(x+y\right)}{x+y}\)

\(=\frac{\left(x+y\right)\left(x^2+y^2\right)+\left(x+y\right)\left(x^2-2xy+y^2\right)}{x+y}\)

\(=\frac{\left(x+y\right)\left\lbrack\left(x^2+y^2\right)+\left(x-y\right)^2\right\rbrack}{x+y}\)

\(=x^2+y^2+\left(x-y\right)^2\)

\(\ge x^2+y^2\) ( đpcm )

( Dấu "=" xảy ra khi \(x=y\) )

+, Do \(x+y=1\) nên \(\left(x+y\right)^2=x+y=1\)

+, \(\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\)

\(=\frac{\left(x+1\right)\left(y+1\right)}{xy}-9+9\)

\(=\frac{xy+x+y+1-9xy}{xy}+9\)

\(=\frac{2-8xy}{xy}+9\)

\(=\frac{2\left(1-4xy\right)}{xy}+9\)

\(=\frac{2\left\lbrack\left(x+y\right)^2-4xy\right\rbrack}{xy}+9\)

\(=\frac{2\left(x-y\right)^2}{xy}+9\)

\(\ge9\) ( đpcm )

( Dấu "=" xảy ra khi \(x=y=\frac12\) )

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+1\)

\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)

\(=\left(x^2-5x\right)^2+10\left(x^2-5x\right)+25\)

\(=\left(x^2-5x+5\right)^2\)

\(=\left\lbrack\left(x^2-5x+\frac{25}{4}\right)-\frac54\right\rbrack^2\)

\(=\left\lbrack\left(x-\frac52\right)^2-\left(\frac{\sqrt5}{2}\right)^2\right\rbrack^2\)

\(=\left(x-\frac52+\frac{\sqrt5}{2}\right)^2\left(x-\frac52-\frac{\sqrt5}{2}\right)^2\)

\(\ge0\) ( đpcm )

( Dấu "=" xảy ra khi \(x\in\left\lbrace\frac{5-\sqrt5}{2};\frac{5+\sqrt5}{2}\right\rbrace\) )

\(4x^8-2x^7+x^6-3x^4+x^2-x+1\)

\(=\left(x^8-2x^7+x^6\right)+\left(3x^8-3x^4+\frac34\right)+\left(x^2-x+\frac14\right)\)

\(=x^6\left(x-1\right)^2+3\left(x^4-\frac12\right)^2+\left(x-\frac12\right)^2\)

\(>0\) ( đpcm )

+, Ta có: \(a+b+c+ab+bc+ac=6abc\) nên \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)

+, \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(=\frac13\left(\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}\right)\)

\(=\frac13\left\lbrack\left(\frac{1}{a^2}+1\right)+\left(\frac{1}{b^2}+1\right)+\left(\frac{1}{c^2}+1\right)-3+\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}+\frac{1}{c^2}\right)+\left(\frac{1}{a^2}+\frac{1}{c^2}\right)\right\rbrack\)

\(\ge\frac13\left(2\sqrt{\frac{1}{a^2}.1}+2\sqrt{\frac{1}{b^2}.1}+2\sqrt{\frac{1}{c^2}+1}-3+2\sqrt{\frac{1}{a^2b^2}}+2\sqrt{\frac{1}{b^2c^2}}+2\sqrt{\frac{1}{a^2c^2}}\right)\)

\(=\frac13\left(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}-3+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\right)\)

\(=\frac23\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)-1\)

\(=\frac{2.6}{3}-1\)

\(=3\) ( đpcm )

( Dấu "=" xảy ra khi \(a=b=c=1\) )

\(x^2+y^2+xy-3x-3y+3\)

\(=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(xy-x-y+1\right)\)

\(=\left(x-1\right)^2+\left(y-1\right)^2+\left\lbrack x\left(y-1\right)-\left(y-1\right)\right\rbrack\)

\(=\left(x-1\right)^2+\left(y-1\right)^2+\left(x-1\right)\left(y-1\right)\)

\(=\left\lbrack\left(x-1\right)^2+\left(x-1\right)\left(y-1\right)+\frac{\left(y-1\right)^2}{4}\right\rbrack+\frac{3\left(y-1\right)^2}{4}\)

\(=\left\lbrack\left(x-1\right)+\frac{y-1}{4}\right\rbrack^2+\frac{3\left(y-1\right)^2}{4}\)

\(\ge0\) ( đpcm )

( Dấu "=" xảy ra khi \(x=y=1\) )

1)

\(a^2-ab+b^2\)

\(=a^2-ab+\frac{b^2}{4}+\frac{3b^2}{4}\)

\(=\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}\)

\(\ge0\) ( đpcm )

( Dấu "=" xảy ra khi \(a=b=0\) )

2)

\(a^2-ab+b^2\)

\(=\left(\frac{3a^2}{4}-\frac{3ab}{2}+\frac{3b^2}{4}\right)+\left(\frac{a^2}{4}+\frac{ab}{2}+\frac{b^2}{4}\right)\)

\(=\frac{3\left(a-b\right)^2}{4}+\frac{\left(a+b\right)^2}{4}\)

\(\ge\frac14\left(a+b\right)^2\) ( đpcm )

( Dấu "=" xảy ra khi \(a=b\) )

\(\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{a^2-ac+c^2}\)

\(=\sum_{cyc}^{a,b,c}\sqrt{a^2-ab+b^2}\)

\(=\sum_{cyc}^{a,b,c}\sqrt{\left(\frac{3a^2}{4}+\frac{3b^2}{4}-\frac{3ab}{2}\right)+\left(\frac{a^2}{4}+\frac{b^2}{4}+\frac{ab}{2}\right)}\)

\(=\sum_{cyc}^{a,b,c}\sqrt{\frac34\left(a-b\right)^2+\frac14\left(a+b\right)^2}\)

\(\ge\sum_{cyc}^{a,b,c}\sqrt{\frac14\left(a+b\right)^2}\)

\(=\sum_{cyc}^{a,b,c}\frac{a+b}{2}\)

\(=\frac{a+b}{2}+\frac{b+c}{2}+\frac{a+c}{2}\)

\(=a+b+c\)

\(=3\) ( đpcm ).

( Dấu "=" xảy ra khi \(a=b=c=1\) )