Chưng minh rằng:
$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\geq \frac{a+b+c}{\sqrt{2}}$
Chưng minh rằng:
$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\geq \frac{a+b+c}{\sqrt{2}}$
Chưng minh rằng:
$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\geq \frac{a+b+c}{\sqrt{2}}$
Theo BĐT Holder thì:
$$\left ( \frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \right )^{2}\left ( a\left ( b+c \right )+b(a+c)+c(a+b) \right )\geqslant (a+b+c)^{3}$$
Do đó ta chỉ cần chứng minh:
$$\sqrt{\frac{(a+b+c)^{3}}{2(ab+bc+ca)}}\geqslant \frac{a+b+c}{\sqrt{2}}\Leftrightarrow a+b+c\geqslant ab+bc+ca$$
Mặt khác,ta có theo AM-GM:
$$4=a+b+c+abc \leqslant (a+b+c)+\frac{(a+b+c)^{3}}{27}\Leftrightarrow (x-3)(x^{2}+3x+36)\geqslant 0\Leftrightarrow x\geqslant 3$$
Trong đó $x=a+b+c>0$.
Ta có:
$4\left ( a+b+c \right )=(a+b+c+abc)(a+b+c)=2(ab+bc+ca)+\left ( a^{2}+b^{2}+c^{2} \right )+abc(a+b+c)$
$\geqslant 2(ab+bc+ca)+(a^{2}+b^{2}+c^{2})+\frac{9abc}{a+b+c}$
Lại có theo BĐT Schur thì:
$$a^{2}+b^{2}+c^{2}+\frac{9abc}{a+b+c}\geqslant 2(ab+bc+ca)$$
Do đó:
$$4\left ( a+b+c \right )\geqslant 2(ab+bc+ca)+2(ab+bc+ca)\Leftrightarrow a+b+c \geqslant ab+bc+ca$$
Vậy ta có đpcm.
0 thành viên, 0 khách, 0 thành viên ẩn danh