Từ giả thiết  a ≤ 1 , b ≤ 1 , c ≤ 1 ta có  a 4 ≤ a 2 , b 6 ≤ b 2 , c 8 ≤ c 2 . Từ đó  a 4 + b 6 + c 8 ≤ a 2 + b 2 + c 2

Lại có:  a − 1 b − 1 c − 1 ≤ 0   v à   a + 1 b + 1 c + 1 ≥ 0 nên

a + 1 b + 1 c + 1 − a − 1 b − 1 c − 1 ≥ 0 ⇔ 2 a b + 2 b c + 2 c a + 2 ≥ 0 ⇔ − 2 a b + b c + c a ≤ 2

Hơn nữa  a + b + c = 0 ⇔ a 2 + b 2 + c 2 = − a b + b c + c a ≤ 2

⇒ a 4 + b 6 + c 8 ≤ 2

Xét \(VT=a+2b+c=1+b\left(1\right)\)

Áp dụng BĐT AG-GM:

\(4\left(1-a\right)\left(1-c\right)\le\left(1-a+1-c\right)^2=\left(2-a-c\right)^2=\left(1+a+b+c-a-c\right)^2=\left(1+b\right)^2\left(2\right)\)

\(\Rightarrow4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(1-b\right)\left(1+b\right)^2\)

Mà \(\left(1-b\right)\left(1+b\right)^2-\left(1-b\right)=\left(1+b\right)\left(1-b^2-1\right)=-b^2\left(1+b\right)\le0,\forall b\ge0\)

Do đó \(\left(1-b\right)\left(1+b\right)^2\le1+b\left(3\right)\)

Từ \(\left(1\right)\left(2\right)\left(3\right)\) ta có ĐPCM

Dấu "=" \(\Leftrightarrow a=c=\dfrac{1}{2};b=0\) 

;

Ta có \(a+b+c=abc\Leftrightarrow\dfrac{a+b+c}{abc}=1\) \(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

Lại có \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(\Leftrightarrow2^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) (đpcm)

Giúp vs mn ơi

Cái cuối là c(1/a+1/b) nha mn

 ≥ 9(a + b + c)

(a2+b2+c2)3(a2+b2+c2)3 ≥ 9(a + b + c)

Lời giải:
Áp dụng BĐT Bunhiacopxky:

$(a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2$

$\Rightarrow a^2+b^2+c^2\geq \frac{(a+b+c)^2}{3}$

$\Rightarrow (a^2+b^2+c^2)^3\geq \frac{(a+b+c)^6}{27}$

Áp dụng BĐT Cô-si: $a+b+c\geq 3\sqrt[3]{abc}=3$

$\Rightarrow (a^2+b^2+c^2)^3\geq \frac{(a+b+c)^6}{27}\geq \frac{(a+b+c).3^5}{27}=9(a+b+c)$
Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=1$