Cho x,y>0 và x+y 2 Tìm GTNN : P = `1/(x^2+y^2) + 3/(xy) + xy`

Đáp án:

 \(Min _P=\dfrac92.\Rightarrow x=y=1.\)

Giải thích các bước giải:

 Ta có:

\(P=\dfrac{1}{x^{2}+y^{2}}+\dfrac3{xy}+xy\\\Rightarrow P=\dfrac{1}{x^{2}+y^{2}}+\dfrac{1}{2xy}+\dfrac{5}{2xy}+\dfrac{5xy}2-\dfrac32xy\\\Rightarrow P\buildrel{{Bunhiacopski}}\over\ge \dfrac{(1+1)^{2}}{x^{2}+2xy+y^{2}}+\dfrac{5}{2xy}+\dfrac{5xy}{2}-\dfrac32xy\\\Rightarrow P\buildrel{{AM-GM}}\over\ge\dfrac{4}{(x+y)^{2}}+2.\sqrt{\dfrac{5}{2xy}.\dfrac{5xy}{2}}-\dfrac32xy\\\Rightarrow P\ge\dfrac4{2^{2}}+5-\dfrac32xy\\\Rightarrow P\ge 1+5-\dfrac32 xy=6-\dfrac32 xy\\\text{Vì:}\\x+y\le 2\\\Rightarrow 2\ge x+y\ge2\sqrt{xy}\\\Rightarrow 2\ge2\sqrt{xy}\\\rightarrow 1\ge xy.\\\Rightarrow \dfrac32 xy\le \dfrac32\\\Rightarrow -\dfrac32 xy\ge -\dfrac32\\\Rightarrow P\ge6-\dfrac32=\dfrac92.\\\text{Dấu "=" xảy ra khi:x=y=1}.\)

`@nguyen``nam500#hoidap247`.