Chủ đề này có 2 trả lời

[kiencoam]

Cho a, b, c >0 và a+b+c=3

Cmr: $\frac{a}{b^{3}+ab}+\frac{b}{c^{3}+bc}+\frac{c}{a^{3}+ca}\geq \frac{3}{2}$

[phongmaths]

Ta có:

BĐT cần CM $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-(\frac{a}{b^{3}+ab}+\frac{b}{c^{3}+bc}+\frac{c}{a^{3}+ca})\leq (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-\frac{3}{2}$

Mà $VT = \sum (\frac{1}{b}-\frac{a}{b^{3}+ab})=\sum (\frac{b}{b^{2}+a})\leq \sum (\frac{b}{2b\sqrt{a}})=\frac{1}{2}\sum (\frac{1}{\sqrt{a}})$

$VP=\sum (\frac{1}{a})-\frac{3}{2}\geq \frac{1}{3}\sum (\frac{1}{\sqrt{a}})^{2}-\frac{3}{2}$

Đặt $\sum (\frac{1}{\sqrt{a}})=t$

Mà $t=\sum (\frac{1}{\sqrt{a}})\geq \frac{9}{\sum \sqrt{a}}\geq \frac{9}{\sqrt{3(a+b+c)}}=3$

BĐT cần CM $t^{2}-\frac{3}{2}\geq \frac{t}{2}\Leftrightarrow \frac{(t-3)(2t+3)}{6}\geq 0$ (BĐT đúng với mọi $t\geq 3$)

Dấu = xãy ra khi a=b=c=1

[MrDat]

Xét $\frac{a}{b^{3}+ab}=\frac{a}{b(a+b^2)}=\frac{a+b^{2}-b^{2}}{b(a+b^{2})}=\frac{1}{b}-\frac{b}{a+b^{2}}$

Có $a+b^{2}\geq 2\sqrt{a}.b$ suy ra $\frac{1}{b}-\frac{b}{a+b^{2}}\geq \frac{1}{b}-\frac{b}{2\sqrt{a}.b}=\frac{1}{b}-\frac{1}{2\sqrt{a}}$

Tương tự $\frac{b}{c^{3}+bc}\geq \frac{1}{c}-\frac{1}{2\sqrt{b}}$

                $\frac{c}{a^{3}+ca}\geq \frac{1}{a}-\frac{1}{2\sqrt{c}}$   

Ta có VT $\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{2}.(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})$

Lại có $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$ nên $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3$

Đặt $\frac{1}{a}=x$ , $\frac{1}{b}=y$ , $\frac{1}{c}=z$

     suy ra VT $\geq x+y+z-\frac{1}{2}(\sqrt{x}+\sqrt{y}+\sqrt{z})$

    Có $\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \frac{x+1+y+1+z+1}{2}$

nên VT$\geq \frac{4.(x+y+z)}{4}-\frac{x+y+z+3}{4}=\frac{3(x+y+z)-3}{4}\geq \frac{2.(x+y+z)}{4}\geq \frac{3}{2}$ (do x+y+z$\geq 3$)

BĐT được CM , dấu bằng xảy ra khi x=y=z=1