ta luôn có \(\left(a+2c\right)^2\ge0\Leftrightarrow a^2+4c^2\ge-4ac\left(1\right)\)
theo bđt cauchy ta có
\(\left\{{}\begin{matrix}a^2+4b^2\ge4ab\left(2\right)\\4b^2+4c^2\ge8bc\left(3\right)\end{matrix}\right.\)
lấy (1)+(2)+(3)\(\Leftrightarrow2a^2+8b^2+8c^2\ge4ab-4ac+8bc\)
\(\Leftrightarrow2(a^2+4b^2+4c^2)\ge4(ab-ac+2bc)\)
rút 2 đi \(\Rightarrow dpcm\)
Chứng minh các BĐT sau:
a/ \(2\left(a^4+1\right)+\left(b^2+1\right)^2\ge2\left(ab+1\right)^2\)
b/ \(3\left(a^2+b^2\right)-ab+4\ge2\left(a\sqrt{b^2+1}+b\sqrt{a^2+1}\right)\)
Cho ba số thực dương a, b, c. Chứng minh rằng:
a) \(\left(a+\frac{4b}{c^2}\right)\left(b+\frac{4c}{a^2}\right)\left(c+\frac{4a}{b^2}\right)\ge64\)
b) \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca\)
Cho a,b,c,d >0 thỏa abcd=1.Chứng minh BĐT sau với \(k\ge2\):
\(\sum\dfrac{1}{\left(1+a\right)^k}\ge\dfrac{4}{2^k}\)
CM BĐT sau
a/ \(\left(a^2-b^2\right)\left(c^2-d^2\right)\le\left(ac-bd\right)^2\) \(\forall a,b,c,d\)
b/ \(\left(1+a^2\right)\left(1+b^2\right)\ge\left(1+ab\right)^2\) \(\forall a,b\)
c/ \(a^2+b^2+1\ge ab+a+b\) \(\forall a,b\)
CM BĐT
a/ \(2a^2+b^2+c^2\ge2a\left(b+c\right)\) \(\forall a,b\)
b/ \(a^2+2b^2+12\ge2b\left(3-a\right)\) \(\forall a,b\)
c/ \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3\) \(\forall a,b\)
_Cho \(\left\{{}\begin{matrix}a^2+b^2+2a-4b+4=0\\c^2+d^2-4c+4d+4=0\end{matrix}\right.\) . Tìm min, max : P = \(\left(a-c\right)^2+\left(b-d\right)^2\) ?
cho a,b>0. chứng minh \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
áp dụng chứng minh bđt sau:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)vớia,b,c>0\)
Cho a > b > 0. Chứng minh rằng:
\(a+\frac{1}{b\left(a-b\right)^2}\ge2\sqrt{2}\)
Chứng minh BĐT \(\sqrt[3]{\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)}\le\dfrac{\left(x+y+z\right)^2}{3}+1\)
với x,y,z>0 và \(Min\left\{xy,yz,zx\right\}\ge1\)
