a, \(\dfrac{a^2+2ab+b^2}{4}\ge ab\)

\(\Leftrightarrow\)a^2+2ab+b^2>=4ab

\(\Leftrightarrow\)a^2-2ab+b^2>=0

\(\Leftrightarrow\)(a-b)^2>=0 (luôn đúng)

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

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

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

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

a: (sina+cosa)^2

=sin^2a+cos^2a+2*sina*cosa

=1+sin2a

b: \(cos^4a-sin^4a=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)\)

\(=cos^2a-sin^2a=cos2a\)

\(a^2+b^2\ge\frac{1}{2}\)

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

\(\Leftrightarrow2a^2+2b^2\ge\left(a+b\right)^2\)

\(\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\) đúng

Vậy ta có đpcm

Không chắc là đúng đâu nhé :D

\(a^2+b^2\ge\frac{1}{2}\)

\(\Leftrightarrow a^2+b^2-\frac{a+b}{2}\ge0\)

\(\Leftrightarrow2a^2+2b^2-a-b\ge0\)

\(\Leftrightarrow2a\left(a+b\right)-\left(a+b\right)\ge0\)

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

\(\Leftrightarrow2a-1\ge0\)

\(\Leftrightarrow a\ge\frac{1}{2}\)

\(\Rightarrow a^2+b^2\ge\frac{1}{2}\)

Áp dụng BĐT bunhiacopxki ta có:

\(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow2.\left(a^2+b^2\right)\ge1\)

\(\Leftrightarrow a^2+b^2\ge\frac{1}{2}\)

Dấu " = " xảy ra <=> a=b=0,5

Ta có :

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

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

\(\Leftrightarrow2a^2+2b^2-2ab-2a-2b+2\ge0\)

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

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

a) Ta có: \(\left(a-1\right)^2\ge0\forall a\)

\(\Leftrightarrow a^2-2a+1\ge0\forall a\)

\(\Leftrightarrow a^2+2a+1\ge4a\forall a\)

\(\Leftrightarrow\left(a+1\right)^2\ge4a\)(đpcm)

b) Áp dụng bất đẳng thức Cosi ta có:

\(a+1\ge2\sqrt{a};b+1\ge2\sqrt{b};c+1\ge2\sqrt{c}\\ \Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)

Dấu = xảy ra khi và chỉ khi a=b=c=1