9.

\(\Leftrightarrow a^2+a^2b^2+b^2+b^2c^2+c^2+c^2a^2\ge6abc\)

\(\Leftrightarrow\left(a^2-2abc+b^2c^2\right)+\left(b^2-2abc+c^2a^2\right)+\left(c^2-2abc+a^2b^2\right)\ge0\)

\(\Leftrightarrow\left(a-bc\right)^2+\left(b-ca\right)^2+\left(c-ab\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;0\right);\left(1;1;1\right);\left(1;-1;-1\right)\) và các hoán vị

10.

\(a^2+b^2+c^2=1\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=1+2\left(ab+bc+ca\right)\)

\(\Leftrightarrow\left(a+b+c\right)^2=1+2\left(ab+bc+ca\right)\)

\(\Rightarrow1+2\left(ab+bc+ca\right)\ge0\Rightarrow ab+bc+ca\ge-\dfrac{1}{2}\)

Lại có:

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow ab+bc+ca\le1\)

11.

Do \(a^2+b^2+c^2=1\Rightarrow\left\{{}\begin{matrix}\left|a\right|\le1\\\left|b\right|\le1\\\left|c\right|\le1\end{matrix}\right.\) \(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge0\)

Do đó:

\(abc+2\left(1+a+b+c+ab+bc+ca\right)\)

\(=1+a+b+c+ab+bc+ca+\left(1+a+b+c+ab+bc+ca+abc\right)\)

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

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

\(=\dfrac{1}{2}\left(a+b+c+1\right)^2+\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge0\) (đpcm)

12.

\(a^4+3\ge4a\)

\(\Leftrightarrow a^4-2a^3+a^2+\left(2a^3-4a^2+2a\right)+\left(3a^2-6a+3\right)\ge0\)

\(\Leftrightarrow a^2\left(a-1\right)^2+2a\left(a-1\right)^2+3\left(a-1\right)^2\ge0\)

\(\Leftrightarrow\left(a^2+2a+3\right)\left(a-1\right)^2\ge0\)

\(\Leftrightarrow\left[\left(a+1\right)^2+2\right]\left(a-1\right)^2\ge0\) (luôn đúng)

14.

\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-3\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+2\ge0\)

\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}-1\right)\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)\ge0\)

\(\Leftrightarrow\left(\dfrac{a^2+b^2-ab}{ab}\right)\left(\dfrac{a^2+b^2-2ab}{ab}\right)\ge0\)

\(\Leftrightarrow\dfrac{\left[\left(a-\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\right]\left(a-b\right)^2}{\left(ab\right)^2}\ge0\) (luôn đúng)

15.

\(\Leftrightarrow x^8+y^8\ge x^6y^2+x^2y^6\)

\(\Leftrightarrow x^8-x^6y^2+y^8-x^2y^6\ge0\)

\(\Leftrightarrow x^6\left(x^2-y^2\right)-y^6\left(x^2-y^2\right)\ge0\)

\(\Leftrightarrow\left(x^6-y^6\right)\left(x^2-y^2\right)\ge0\)

\(\Leftrightarrow\left(x^2-y^2\right)^2\left(x^4+x^2y^2+y^4\right)\ge0\) (luôn đúng)

13.

\(\Leftrightarrow\dfrac{y\left(x+z\right)}{xz}+\dfrac{x+z}{y}\le\dfrac{\left(x+z\right)^2}{xz}\)

\(\Leftrightarrow\dfrac{y}{zx}+\dfrac{1}{y}\le\dfrac{x+z}{xz}\)

\(\Leftrightarrow y^2+zx\le y\left(x+z\right)\)

\(\Leftrightarrow xy+yz\ge y^2+zx\)

\(\Leftrightarrow xy-y^2+yz-zx\ge0\)

\(\Leftrightarrow y\left(x-y\right)-z\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)\left(y-z\right)\ge0\) (hiển nhiên đúng do \(x\ge y\ge z>0\))

16.

\(\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\ge abc\left(a+b+c\right)\)

\(\Leftrightarrow2\left(ab\right)^2+2\left(bc\right)^2+2\left(ca\right)^2\ge2abc\left(a+b+c\right)\)

\(\Leftrightarrow\left(a^2b^2-2a^2bc+a^2c^2\right)+\left(a^2c^2-2abc^2+b^2c^2\right)+\left(b^2c^2-2ab^2c+c^2a^2\right)\ge0\)

\(\Leftrightarrow\left(ab-ac\right)^2+\left(ac-bc\right)^2+\left(bc-ca\right)^2\ge0\) (luôn đúng)

cái khúc này làm sao suy ra được vậy

Rút gọn x+z (dương) là được em

Anh ơi em vẫn chưa hiểu anh giải thích rõ hơn dc ko

Tử số 2 vế dòng đầu đều có x+z, rút gọn nó đi là được dòng 2 thôi

Anh ơi dòng hai là sao v

Đặt \(\dfrac{a}{b}+\dfrac{b}{a}=x\) rồi bạn phân tích thành nhân tử sẽ thấy tại sao lại như vậy