Bất đẳng thức $20$

Bài Toán:Cho các số thực dương a,b,c thay đổi thỏa mãn điều kiện $a^{2}+b^{2}+c^{2}=14$Tìm GTLN của biểu thức$$P= \dfrac{4\left ( a+c \right )}{a^{2}+3c^{2}+28} + \dfrac{4a}{a^{2}+bc+7} - \dfrac{5}{\left ( a+b \right )^{2}}-\dfrac{3}{a\left ( b+c \right )}$$

Lời giải:Ta sẽ CM : $P\leq \dfrac{8}{15}$

 Ta có : $(a+c)^2=(a+c\sqrt{3}.\dfrac{1}{\sqrt{3}})^2\leq (a^2+3c^2)(1+\dfrac{1}{3})$

$= > a^2+3c^2\geq \dfrac{3(a+c)^2}{4}$

$= > \dfrac{ 4(a+c)}{a^2+3c^2+28}$

$\leq \dfrac{4(a+c)}{\dfrac{3(a+c)^2}{4}+28}$

$=\dfrac{16(a+c)}{3(a+c)^2+112}$

$=\dfrac{16(a+c)}{3\left [ (a+c)^2+16 \right ]+64}$

$\leq \dfrac{16(a+c)}{3.2.\sqrt{16(a+c)^2}+64}$

$=\dfrac{2(a+c)}{3(a+c)+8}$  (1)

$\dfrac{4a}{a^2+bc+7}=\dfrac{8a}{2a^2+2bc+14}$

$=\dfrac{8a}{2a^2+2bc+a^2+b^2+c^2}$

$=\dfrac{8a}{3a^2+(b+c)^2}$

$\leq \dfrac{8a}{2a^2+2a(b+c)}$

$=\dfrac{4}{a+b+c}$  (2)

 $\dfrac{3}{a(b+c)}\geq \dfrac{3}{\dfrac{(a+b+c)^2}{4}}=\dfrac{12}{(a+b+c)^2}=\dfrac{1}{3}(\dfrac{6}{a+b+c}-1)^2+\dfrac{4}{a+b+c}-\dfrac{1}{3}\geq \dfrac{4}{a+b+c}-\dfrac{1}{3}$

$= > -\dfrac{3}{a(b+c)}\leq \dfrac{1}{3}-\dfrac{4}{a+b+c}$    (3)

$\dfrac{5}{(a+b)^2}+\dfrac{1}{5}\geq 2\sqrt{\dfrac{5}{(a+b)^2}.\dfrac{1}{5}}=\dfrac{2}{a+b}$

$= > -\dfrac{5}{(a+b)^2}\leq \dfrac{1}{5}-\dfrac{2}{a+b}$   (4)

    Cộng theo vế (1) ,(2),(3) (4)

$= > P\leq \dfrac{2(a+c)}{3(a+c)+8}+\dfrac{4}{a+b+c}-\dfrac{4}{a+b+c}+\dfrac{1}{3}-\dfrac{2}{a+b}+\dfrac{1}{5}\leq \dfrac{5}{18}$

$< = > \dfrac{a+c}{3(a+c)+8}-\dfrac{1}{a+b}\leq 0$

$< = > (a+c)(a+b)-(3(a+c)+8)\leq 0$

$< = > (a+c)(a+b-3)\leq 8$

$< = > (a+c)(2a+2b-6)\leq 16$

Theo AM-GM có :

$(a+c)(2a+2b-6)\leq \dfrac{(3a+2b+c-6)^2}{4}\leq 16$

$< = > (3a+2b+c)^2-12(3a+2b+c)-28\leq 0$

$< = > (3a+2b+c-14)(3a+2b+c+2)\leq 0$

$< = > 3a+2b+c\leq 14$

Nhưng BĐT này luôn đúng vì $3a+2b+c\leq \sqrt{(a^2+b^2+c^2)(3^2+2^2+1^2)}=14$

   Do đó ta có ĐPCM .

 Dấu  =  xảy ra tại $a=3,b=2,c=1$