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
