BẤT ĐẲNG THỨC 4

Cho $a,b,c>0$ và $16(a+b+c)\geq \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$.

Chứng minh rằng:$\dfrac{1}{\left [ a+b+\sqrt{2(a+c)} \right ]^3}+\dfrac{1}{\left [ b+c+\sqrt{2(a+b)} \right ]^3}+\dfrac{1}{\left [ a+c+\sqrt{2(b+c)} \right ]^3}\leq \dfrac{8}{9}$

Lời Giải:

Theo BĐT Cosi cho 3 số ta có :

 $\sum \dfrac{1}{\left [ a+b+\sqrt{2(a+c)} \right ]^3}=\sum \dfrac{1}{\left [ (a+b)+\dfrac{\sqrt{a+c}}{\sqrt{2}}+\dfrac{\sqrt{a+c}}{\sqrt{2}} \right ]^3}$

$\leq \sum \dfrac{1}{27.(a+b).\dfrac{\sqrt{a+c}}{\sqrt{2}}.\dfrac{\sqrt{a+c}}{\sqrt{2}}}=\frac{2}{27}\sum \dfrac{1}{(a+b)(a+c)}=\dfrac{2}{27}.\dfrac{\sum (b+c)}{(a+b)(b+c)(c+a)}$

$=\dfrac{4(\sum a)}{27(a+b)(b+c)(c+a)}$  (1)

 Mặt khác ta lại có : $(a+b)(b+c)(c+a)\geq \dfrac{8(\sum a)(\sum ab)}{9}$ (2)

 Từ (1),(2) $= > \sum \dfrac{1}{\left [ a+b+\sqrt{2(a+b)} \right ]^3}\leq \dfrac{4(\sum a)}{27.\dfrac{8(\sum a)(\sum ab)}{9}}=\dfrac{1}{6(\sum ab)}$  (3)

 Ta có : $16(\sum a)\geq \sum \dfrac{1}{a}=\dfrac{\sum ab}{abc}= > \sum ab\leq 16abc(\sum a)\leq 16.\dfrac{(\sum ab)^2}{3}= > \sum ab\geq \dfrac{3}{16}$  (4)

 Từ (3),(4) $= >\sum \dfrac{1}{\left [ a+b+\sqrt{2(a+c)} \right ]^3}\leq \dfrac{1}{6.\dfrac{3}{16}}=\dfrac{8}{9}$

 Do đó ta có ĐPCM.  Dấu = xảy ra khi $< = > \left\{\begin{matrix} a+b=\dfrac{\sqrt{a+c}}{\sqrt{2}} & & & \\ b+c=\dfrac{\sqrt{a+b}}{\sqrt{2}} & & & \\ a+c=\dfrac{\sqrt{b+c}}{\sqrt{2}} & & & \\ a=b=c=\dfrac{1}{4} & & & \end{matrix}\right.< = > a=b=c=\dfrac{1}{4}$