thien huu

Ta có: $(\sum \sqrt{\frac{a}{a^2+b+c}.\frac{a}{3}})^2\leq (\sum \frac{a}{a^2+b+c}).\frac{a+b+c}{3}$

Mặt khác: $\sum \frac{a}{a^2+b+c}=\sum \frac{a(1+b+c)}{(a^2+b+c)(1+b+c)}\leq \frac{a+b+c+2(ab+bc+ca)}{(a+b+c)^2}$

$3(a^2+b^2+c^2)\geq (a+b+c)^2<=>a+b+c \leq 3= a^2+b^2+c^2$

Do đó $\sum \frac{a}{a^2+b+c}\leq \frac{a^2+b^2+c^2+2(ab+bc+ca)}{(a+b+c)^2}=\frac{(a+b+c)^2}{(a+b+c)^2}= 1$

Suy ra $(\sum \sqrt{\frac{a}{a^2+b+c}.\frac{a}{3}})^2\leq \frac{a+b+c}{3}\leq 1$

<=>$\sum \sqrt{\frac{a^2}{3(a^2+b+c)}}\leq 1<=>\sum \sqrt{\frac{a^2}{a^2+b+c}}\leq \sqrt{3}$

Dấu "=" xảy ra <=> a=b=c=1

Lời giải bài bất đẳng thức

Từ giả thiết, ta có:

$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}= \frac{3}{2}$

$<=>\frac{3}{2}+\sum \frac{1}{1+a}= 3<=>\frac{3}{2}=3-\sum \frac{1}{1+a}$

$<=>\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}= \frac{3}{2}$

Ta có nhận xét: $\sqrt{\frac{a^2+1}{2}}= \sqrt{\frac{(1+1)(a^2+1)}{4}}\geq \frac{\sqrt{(a+1)^2}}{2}= \frac{a+1}{2}$

Tương tự: $\sqrt{\frac{b^2+1}{2}}\geq\frac{b+1}{2} ,\sqrt{\frac{c^2+1}{2}}\geq\frac{c+1}{2}$

VT$\geq \sum \frac{a+1}{2}+3= \sum \frac{a+1}{2}+2\sum \frac{a}{a+1}$

$\sum \frac{a+1}{2}+\sum \frac{2a}{a+1}\geq \sum 2\sqrt{\frac{a+1}{2}.\frac{2a}{a+1}}= 2\sum \sqrt{a}$=VP

Dấu "=" xảy ra <=>a=b=c=1

$abc(a+b)(b+c)(c+a)=(bc+ca)(ca+ab)(ab+bc)\leq \frac{(2(ab+bc+ca))^3}{27}$

Mặt khác $ab+bc+ca\leq \frac{(a+b+c)^2}{3}= \frac{1}{3}$

Suy ra VT$\leq \frac{8}{729}$

Đặt $x=a+c, y=b+c$

Suy ra $xy=1$ và $x-y=(a+c)-(b+c)=a-b$

VT=$\frac{1}{(x-y)^2}+\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{(x-y)^2}+\frac{x^2+y^2}{x^2y^2}= \frac{1}{(x-y)^2}+x^2+y^2$

=$\frac{1}{(x-y)^2}+(x-y)^2+2\geq 2+2= 4$=VP

Hệ thức Vi-ét: $x_1+x_2=-\frac{b}{a}$ và $x_1x_2=\frac{c}{a}$

Max

M=$\frac{(a-b)(2a-c)}{a(a-b+c)}= (1-\frac{b}{a})\frac{2a-c}{a-b+c}= (1-\frac{b}{a})\frac{\frac{2a-c}{a}}{\frac{a-b+c}{a}}$(Vì $a\neq 0$)

$M=(1-\frac{b}{a})\frac{2-\frac{c}{a}}{1-\frac{b}{a}+\frac{c}{a}}= \frac{(1+x_1+x_2)(2-x_1x_2)}{1+x_1+x_2+x_1x_2}= \frac{2(1+x_1+x_2)-x_1x_2(x_1+x_2+1)}{1+x_1+x_2+x_1x_2}$

$= 2-\frac{x_1x_2(x_1+x_2+3)}{1+x_1+x_2+x_1x_2}\leq 2$(do $x_1,x_2\in \left [ 0;1 \right ]$)

Min

Đặt $t=x_1+x_2, u=x_1x_2<=>t^2\geq 4u$

M=$\frac{(t+1)(2-u)}{1+t+u}\geq \frac{(t+1)(8-t^2)}{t^2+4t+4}$

Ta chứng minh $M\geq \frac{3}{4}$. Thật vậy BĐT <=>$(2-t)(4t^2+15t+10)\geq 0$(luôn đúng)