Chủ đề này có 1 trả lời

[zack]

cho a,b,c là các số thuc duong. Tìm GTLN của biểu thuc

$\frac{ab}{(a+b)^{2}}+\frac{ac}{(a+c)^{2}}+\frac{bc}{(c+b)^{2}}-\frac{4abc}{(a+b)(b+c(c+a)}$

[tap lam toan]

Solution:
Ta sẽ chứng minh $P_{max}=\frac{1}{4}$
$$\frac{ab}{\left ( a+b \right )^{2}}+\frac{bc}{\left ( b+c \right )^{2}}+\frac{ca}{\left ( c+a \right )^{2}}-\frac{4abc}{\left ( a+b \right )\left ( b+c \right )\left ( c+a \right )}-\frac{1}{4}$$

$$=\frac{4ab\left ( b+c \right )^{2}\left ( c+a \right )^{2}-\left ( a+b \right )^{2}\left ( b+c \right )^{2}\left ( c+a \right )^{2}+4bc\left ( a+b \right )^{2}\left ( a+c \right )^{2}+4ca\left ( a+b \right )^{2}\left ( b+c \right )^{2}-16abc\left ( a+b \right )\left ( b+c \right )\left (c+a  \right )}{4\left ( a+b \right )^{2}\left ( b+c \right )^{2}\left ( c+a \right )^{2}}$$

$$=\frac{-\left ( a-b \right )^{2}\left ( b+c \right )^{2}\left ( c+a \right )^{2}+4bc\left ( a+b \right )\left ( a+c \right )\left ( a-b \right )\left ( a-c \right )+4ca\left ( b+a \right )\left ( b+c \right )\left ( b-c \right )\left ( b-a \right )}{4\left ( a+b \right )^{2}\left ( b+c \right )^{2}\left ( c+a \right )^{2}}$$

$$=\frac{-\left ( a-b \right )^{2}\left ( b+c \right )^{2}\left ( c+a \right )^{2}+4c\left ( a+b \right )\left ( ab+c^{2} \right )\left ( a-b \right )^{2}}{4\left ( a+b \right )^{2}\left ( b+c \right )^{2}\left ( c+a \right )^{2}}$$

$$=\frac{-\left ( a-b \right )^{2}\left [ \left ( ab+c^{2}+ac+bc \right )^{2}-4\left ( ac+bc \right )\left ( ab+c^{2} \right ) \right ]}{4\left ( a+b \right )^{2}\left ( b+c \right )^{2}\left ( c+a \right )^{2}}=\frac{-\left ( a-b \right )^{2}\left ( b-c \right )^{2}\left ( c-a \right )^{2}}{4\left ( a+b \right )^{2}\left ( b+c \right )^{2}\left ( c+a \right )^{2}} \leq0$$

Đẳng thức xảy ra khi và chỉ khi 2 số bất kì bằng nhau (Q.E.D)