Question

a^4+b^4+c^2+1≥2a(ab\(^2\)-a+c+1)

Answer 1

Lời giải:

Xét hiệu:

$a^4+b^4+c^2+1-2a(ab^2-a+c+1)=a^4+b^4+c^2+1-2a^2b^2+2a^2-2ac-2a$

$=(a^4+b^4-2a^2b^2)+(c^2+a^2-2ac)+(a^2-2a+1)$

$=(a^2-b^2)^2+(c-a)^2+(a-1)^2\geq 0$

$\Rightarrow a^4+b^4+c^2+1\geq 2a(ab^2-a+c+1)$

Ta có đpcm.

Dấu "=" xảy ra khi \(\left\{\begin{matrix} a^2=b^2\\ c=a\\ a=1\end{matrix}\right.\Leftrightarrow \pm b=a=c=1\)

Answer 2

\(VT-VP=\frac{\left(\sqrt{2}a^2-\sqrt{2}b^2+c+1-2a\right)^2}{4}+\frac{\left(\sqrt{2}a^2-\sqrt{2}b^2+2a-c-1\right)^2}{4}+\frac{\left(c-1\right)^2}{2}\ge0\)