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 \(x \geq 1\) thì \(x^{8} - x^{7} + x^{2} - x + 1\)

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

Ta có: \(a2b2+b2c2≥2⋅a2b2⋅b2c2=2⋅a2c2=2⋅acb2a2​+c2b2​≥2⋅b2a2​⋅c2b2​​=2⋅c2a2​​=2⋅ca​\)

\(b2c2+c2a2≥2⋅b2c2⋅c2a2=2⋅b2a2=2⋅bac2b2​+a2c2​≥2⋅c2b2​⋅a2c2​​=2⋅a2b2​​=2⋅ab​\)

\(c2a2+a2b2≥2⋅c2a2⋅a2b2=2⋅c2b2=2⋅cba2c2​+b2a2​≥2⋅a2c2​⋅b2a2​​=2⋅b2c2​​=2⋅bc​\)

Do đó: \((a2b2+b2c2)+(b2c2+c2a2)+(c2a2+a2b2)≥2(ac+ba+cb)(b2a2​+c2b2​)+(c2b2​+a2c2​)+(a2c2​+b2a2​)≥2(ca​+ab​+bc​)\)

=>\(a2b2+b2c2+c2a2≥ac+ba+cbb2a2​+c2b2​+a2c2​≥ca​+ab​+bc​\)

x5+y5>(x2+y2)(x+y) 

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

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

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

\(= \left(\right. x - y \left.\right)^{2} + \left(\right. x^{2} + y^{2} \left.\right) \left(\right. x + y \left.\right) \geq \left(\right. x^{2} + y^{2} \left.\right) \left(\right. x + y \left.\right)\) suy ra
x4−x3y+x2y2−xy3+y4>x2+y2x4−x3y+x2y2−xy3+y4>x2+y2

Ta có \(x+y=1x+y=1\)

\((1+1x)(1+1y)=(1+x+yx)(1+x+yy)=(2+yx)(2+xy)(1+x1​)(1+y1​)=(1+xx+y​)(1+yx+y​)=(2+xy​)(2+yx​)\)

\(=5+2xy+2yx=5+2(xy+yx)=5+y2x​+x2y​=5+2(yx​+xy​)\) \(xy+yx≥2xy.yx=2⇒2(xy+yx)≥4⇒5+2(xy+yx)≥9yx​+xy​≥2yx​.xy​​=2⇒2(yx​+xy​)≥4⇒5+2(yx​+xy​)≥9\)

Dấu ''='' xảy ra khi x = y 

\((x−1)(x−2)(x−3)(x−4)+1≥0⇔[(x−1)(x−4)][(x−2)(x−3)]+1≥0⇔(x2−x−4x+4)(x2−2x−3x+6)+1≥0⇔(x2−5x+4)(x2−5x+6)+1≥0⇔(x2−5x)2+4(x2−5x)+6(x2−5x)+4⋅6+1≥0⇔(x2−5x)2+10(x2−5x)+25≥0⇔(x2−5x)2+2⋅(x2−5x)⋅5+52≥0⇔(x2−5x+5)2≥0(luoˆn đuˊng)\) 

Giả thiết =>\(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{a b} + \frac{1}{b c} + \frac{1}{c a} = 6\). (1)

Ta có \(\left(\right. \frac{1}{a} - 1 \left.\right)^{2} \geq 0\)

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

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

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

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

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

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

Suy ra \(\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} \geq 3\).

Ta có \(x^{2} + y^{2} + x y - 3 x - 3 y + 3\)

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

\(= \left(\right. x - 1 \left.\right)^{2} + \left(\right. y - 1 \left.\right)^{2} + \left(\right. x - 1 \left.\right) \left(\right. y - 1 \left.\right) \geq 0\)                       

(do \(a^{2} + a b + b^{2} = \frac{1}{4} \left(\right. 4 a^{2} + 4 a b + 4 b^{2} \left.\right) = \frac{1}{4} \left(\right. 2 a + b \left.\right)^{2} + \frac{3}{4} b^{2} \geq 0\))

Ta có \(\sqrt{a^{2} - a b + b^{2}} = \sqrt{\frac{1}{4} \left(\right. a + b \left.\right)^{2} + \frac{3}{4} \left(\right. a - b \left.\right)^{2}} \&\text{nbsp}; \geq \frac{1}{2} \left(\right. a + b \left.\right)\).

Đẳng thức xảy ra khi và chỉ khi \(a = b\).

Ta có\(\sqrt{b^{2} - b c + c^{2}} \geq \frac{1}{2} \left(\right. b + c \left.\right)\) và \(\sqrt{c^{2} - c a + c a} \geq \frac{1}{2} \left(\right. c + a \left.\right)\).

=> \(\sqrt{a^{2} - a b + b^{2}} + \sqrt{b^{2} - b c + c^{2}} + \sqrt{c^{2} - c a + a^{2}} \geq \frac{1}{2} \left(\right. a + b + b + c + c + a \left.\right)\)

\(= \left(\right. a + b + c \left.\right) = 3\)                                                                           

Vậy \(\sqrt{a^{2} - a b + b^{2}} + \sqrt{b^{2} - b c + c^{2}} + \sqrt{c^{2} - c a + a^{2}} \geq 3\).

Đẳng thức xảy ra khi và chỉ khi \(a = b = c = \frac{a + b + c}{3} = 1\).

1) \(a2−ab+b2a2−ab+b2\)

\(=(a2−2⋅a⋅12b+14b2)+34b2=(a−12b)2+34b2≥0∀a,b=(a2−2⋅a⋅21​b+41​b2)+43​b2=(a−21​b)2+43​b2≥0∀a,b\)

Dấu "=" xảy ra khi: \({a−12b=0b=0⇔a=b=0{a−21​b=0b=0​⇔a=b=0\) 

2) \(a2−ab+b2≥14(a+b)2a2−ab+b2≥41​(a+b)2\)

\(<=>a2-ab+b2\ge14(a2+2ab+b2)<=>a2-ab+b2\ge14a2+12ab+14b2<=>34a2-32ab+34b2\ge0<=>34(a2-2ab+b2)\ge0<=>34(a-b)2\ge0(luônđúng)<=>a2-ab+b2\ge41(a2+2ab+b2)<=>a2-ab+b2\ge41a2+21ab+41b2<=>43a2-23ab+43b2\ge0<=>43(a2-2ab+b2)\ge0<=>43(a-b)2\ge0(luônđúng)\)

Dấu "=" xảy ra khi: `a-b=0<=>a=b`