Question

Chứng minh rằng:

a) \(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}\ge2\)

b) Cho \(x^2+y^2=4\)

Chứng minh rằng: \(\sqrt{2\left(x+y\right)+6}+\sqrt{22-6\left(x+y\right)}\ge4\sqrt{2}\)

Answer 1

a) Ta có :  \(\left(\sqrt{\sqrt{x^2+x+1}}\right)^2\) ; \(\left(\sqrt{\sqrt{x^2-x+1}}\right)^2\)

ko âm nên áp dụng bđt \(a^2\)+\(b^2\)\(\ge\)2ab

 \(\left(\sqrt{\sqrt{x^2+x+1}}\right)^2\)+\(\left(\sqrt{\sqrt{x^2-x+1}}\right)^2\)\(\ge\)\(2\left(\sqrt[4]{\left(x^2+x+1\right)\left(x^2-x+1\right)}\right)\)

\(\Leftrightarrow\)\(\sqrt{x^2+x+1}\)+\(\sqrt{x^2-x+1}\)\(\ge\)\(2\left(\sqrt[4]{x^4+x+1}\right)\)\(\ge\)\(2\)\(\forall x\)