Question

Chứng minh 1,\(x^4+5>x^2+4x\)

2, Nếu \(a\ge4,b\ge5,c\ge6,a^2+b^2+c^2=90\Rightarrow a+b+c\ge16\)

Answer 1

1/ \(x^4+5>x^2+4x\)

\(\Leftrightarrow x^4-2x^2+1+x^2-4x+4>0\)

\(\Leftrightarrow\left(x^2-1\right)^2+\left(x-2\right)^2>0\) đúng vì đấu = không xảy ra

Answer 2

2/ Ta có:

\(a=\sqrt{90-b^2-c^2}\le\sqrt{90-5^2-6^2}< 6\)

Tương tự: \(\left\{{}\begin{matrix}b< 7\\c\le7\end{matrix}\right.\)

\(\Rightarrow\left(a-4\right)\left(a-9\right)+\left(b-5\right)\left(b-8\right)+\left(c-6\right)\left(c-7\right)\le0\)

\(\Leftrightarrow a^2+b^2+c^2-13\left(a+b+c\right)+118\le0\)

\(\Leftrightarrow a+b+c\ge16\)