1a)\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

b)\(\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)

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

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

2a)\(a^2+\dfrac{b^2}{4}\ge ab\)

\(\Leftrightarrow a^2-ab+\dfrac{b^2}{4}\ge0\)

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

\(\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2\ge0\)(luôn đúng)

b)Đã cm

c)\(a^2+b^2+1\ge ab+a+b\)

\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)

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

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

Dấu bằng xảy ra khi a=b=1

2. a) a² + \(\dfrac{b^2}{4}\)≥ab

<=> a² - ab + \(\dfrac{b^2}{4}\)≥ 0

<=> a² -2.\(\dfrac{b}{2}a+\left(\dfrac{b}{2}\right)^2\) ≥ 0

<=> \(\left(a-\dfrac{b}{2}\right)^2\)≥ 0 ( luôn đúng )

=> đpcm

b) ( a + b)² ≤ 2( a² + b²)

<=> a² + 2ab + b² - 2a² - 2b² ≤ 0

<=> - ( a² - 2ab + b² ) ≤ 0

<=> - ( a - b)² ≤ 0 ( luôn đúng )

=> đpcm

c) a² + b² + 1 ≥ ab + a + b

<=> 2( a² + b² + 1 ) ≥ 2( ab + a + b)

<=> a² - 2ab + b² + a² - 2a + 1 + b² - 2b + 1 ≥ 0

<=> ( a - b)² + ( a - 1)² + ( b - 1)² ≥ 0 ( luôn đúng )

=> đpcm

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

\(\Rightarrow VT=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}-\left(a+b+c\right)\)

Mặt khác áp dụng \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Rightarrow\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge a+b+b+c+a+c=2\left(a+b+c\right)\)

\(\Rightarrow VT\ge2\left(a+b+c\right)-\left(a+b+c\right)=a+b+c\) (đpcm)

Lời giải:
BĐT \(\Leftrightarrow \frac{a^2+b^2+2}{(a^2+1)(b^2+1)}\geq \frac{2}{ab+1}\)

$\Leftrightarrow (a^2+b^2+2)(ab+1)\geq 2(a^2b^2+a^2+b^2+1)$

$\Leftrightarrow a^3b+a^2+ab^3+b^2+2ab+2\geq 2a^2b^2+2a^2+2b^2+2$

$\Leftrightarrow a^3b+ab^3+2ab\geq 2a^2b^2+a^2+b^2$

$\Leftrightarrow ab(a^2+b^2-2ab)-(a^2+b^2-2ab)\geq 0$

$\Leftrightarrow ab(a-b)^2-(a-b)^2\geq 0$

$\Leftrightarrow (a-b)^2(ab-1)\geq 0$

Điều này luôn đúng với mọi $ab\geq 1$ 

Do đó ta có đpcm 

Dấu "=" xảy ra khi $a=b$ hoặc $ab=1$

\(\dfrac{a+b}{2}\ge\sqrt{ab}\)

\(\Rightarrow a+b\ge2\sqrt{ab}\)

\(\Rightarrow a+b-2\sqrt{ab}\ge0\)

\(\Rightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (đúng)

Dấu "=" xảy ra khi: \(a=b\)

\(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)

Chuyển vế và CM tương tự