⚡Nếu \(x<1\) thì \(x^8-x^7+x^2-x+1\)

\(=x^8+x^2\left(\right.1-x^5\left.\right)+\left(\right.1-x\left.\right)>0\).

⚡Nếu \(x\geq1\) thì \(x^8-x^7+x^2-x+1\)

\(=x^7\left(\right.x-1\left.\right)+x\left(\right.x-1\left.\right)+1>0\).

Bất đẳng thức cần chứng minh tương đương với   \(2\left(\right.\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\left.\right)\geq2\left(\right.\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\left.\right)\)

Xét dấu hiệu \(2\left(\right.\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\left.\right)-2\left(\right.\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\left.\right)\)

\(=\left(\right.\frac{a}{b}-\frac{b}{c}\left.\right)^2+\left(\right.\frac{b}{c}-\frac{c}{a}\left.\right)^2+\left(\right.\frac{c}{a}-\frac{a}{b}\left.\right)^2\geq0\)

x5+y5>(x2+y2)(x+y) (1)

Từ giả thiết \(x>\sqrt{2}\) suy ra \(x^2>2\) suy ra \(x^5>2x^3\), từ đó   

\(x^5+y^5>2\left(\right.x^3+y^3\left.\right)\)

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

\(=\left(\right.x-y\left.\right)^2+\left(\right.x^2+y^2\left.\right)\left(\right.x+y\left.\right)\geq\left(\right.x^2+y^2\left.\right)\left(\right.x+y\left.\right)\) suy ra (1), điều phải chứng minh

Chú ý rằng \(x+y=1\) nên \(\left(\right.1+\frac{1}{x}\left.\right)\left(\right.1+\frac{1}{y}\left.\right)-9\)

\(=\frac{\left(\right.x+1\left.\right)\left(\right.y+1\left.\right)-9xy}{xy}=\frac{2-8xy}{xy}\)  

\(=\frac{2\left(\right.1-4xy\left.\right)}{xy}=\frac{2\left(\right.\left(\right.x+y\left.\right)^2-4xy\left.\right)}{xy}\)

\(=\frac{2\left(\right.x-y\left.\right)^2}{xy}\geq0\)

Chú ý rằng \(1 + 4 = 2 + 3\), ta đặt  \(t=\left(\right.x-1\left.\right)\left(\right.x-4\left.\right)=x^2-5x+4\) thì 

\(\left(\right.x-2\left.\right)\left(\right.x-3\left.\right)=x^2-5x+6=t+2\)

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

\(=t\left(\right.t+2\left.\right)+1=t^2+2t+1=\left(\right.t+1\left.\right)^2\geq0\)

Dẳng thức chỉ xảy ra khi \(t=-1\)

hay \(x^2-5x+4=-1\)

\(x^2-5x+5=0\)

\(x=\frac{5 \pm\sqrt{5}}{2}\).

Vế trái bất đẳng thức cần chứng minh là \(x^6\left(\right.x-1\left.\right)^2+3\left(\right.x^4-\frac{1}{2}\left.\right)^2+\left(\right.x-\frac{1}{2}\left.\right)^2\).

Từ đó suy ra đpcm.

Ta có \(\left(\right.\frac{1}{a}-1\left.\right)^2\geq0\)

\(\frac{1}{a^2}+1\geq\frac{2}{a}\) nên 

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq2\left(\right.\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left.\right)-3\) (2)

Lại có \(\frac{1}{a^2}+\frac{1}{b^2}\geq\frac{2}{ab}\) nên 

\(2\left(\right.\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\left.\right)\geq2\left(\right.\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\left.\right)\) (3)

Cộng (2) và (3) theo vế  và sử dụng (1)  ta có    

\(3\left(\right.\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\left.\right)\geq2\left(\right.\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left.\right)-3=2.6-3=9\) 

Suy ra \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq3\).

Ta có \(x^2+y^2+xy-3x-3y+3\)

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

\(=\left(\right.x-1\left.\right)^2+\left(\right.y-1\left.\right)^2+\left(\right.x-1\left.\right)\left(\right.y-1\left.\right)\geq0\)                       

(do \(a^2+ab+b^2=\frac{1}{4}\left(\right.4a^2+4ab+4b^2\left.\right)=\frac{1}{4}\left(\right.2a+b\left.\right)^2+\frac{3}{4}b^2\geq0\))

bất đẳng thức xảy ra khi và chỉ khi a=b=c =1

dấu = xảy ra khi a-b=0<=> a=b