Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\sqrt{\frac{1}{36}\left(11a+7b\right)^2+\frac{59\left(a-b\right)^2}{36}}+\sqrt{\frac{1}{36}\left(7a+11b\right)+\frac{59\left(a-b\right)^2}{36}}\)
\(=\sqrt{\frac{1}{16}\left(3a+5b\right)^2+\frac{5\left(a-b\right)^2}{16}}+\sqrt{\frac{1}{16}\left(5a+3b\right)^2+\frac{5\left(a-b\right)^2}{16}}\)
\(\ge\frac{1}{6}\left(11a+7b\right)+\frac{1}{6}\left(7a+11b\right)+\frac{1}{4}\left(3a+5b\right)+\frac{1}{4}\left(5a+3b\right)\)
\(=5\left(a+b\right)=5.2016=10080\)
Vì xyz=1\(\Rightarrow x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x\sqrt{x}\)
Tương tự \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2=\left(x+y\right)\ge2z\sqrt{z}\)
\(\Rightarrow P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)
Đặt \(x\sqrt{x}+2y\sqrt{y}=a;y\sqrt{y}+2z\sqrt{z}=b;z\sqrt{z}+2x\sqrt{x}=c\)
\(\Rightarrow x\sqrt{x}=\frac{4c+a-2b}{9};y\sqrt{y}=\frac{4a+b-2c}{9};z\sqrt{z}=\frac{4b+c-2a}{9}\)
\(\Rightarrow P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{a}+\frac{4b+c-2a}{b}\right)\)
\(=\frac{2}{9}\text{ }\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\ge\frac{2}{9}\left(4.3+2-6\right)=2\)
Min P =2 khi và chỉ khi a=b=c khi va chỉ khi x=y=z=1
Từ điều kiện suy ra \(\sqrt{xy}+\sqrt{x}+\sqrt{y}\ge3\)
Áp dụng BĐT Cô-si, ta có :
\(3\le\sqrt{xy}+\sqrt{x}.1+\sqrt{y}.1\le\frac{x+y}{2}+\frac{x+1}{2}+\frac{y+1}{2}\)
\(\Rightarrow x+y\ge2\)
Ta có : \(\frac{x^2}{y}+y\ge2\sqrt{\frac{x^2}{y}.y}=2x\); \(\frac{y^2}{x}+x\ge2\sqrt{\frac{y^2}{x}.x}=2y\)
\(\Rightarrow\frac{x^2}{y}+\frac{y^2}{x}+x+y\ge2x+2y\)
\(\Rightarrow P=\frac{x^2}{y}+\frac{y^2}{x}\ge x+y\ge2\)
Vậy GTNN của P là 2 khi x = y = 1
Áp dung BĐT co- si, ta có:
\(y+z\le\sqrt{2\left(y^2+z^2\right)}\)
D đó: \(\frac{x^2}{y+z}\ge\frac{x^2}{\sqrt{2\left(y^2+z^2\right)}}\)
tương tự: \(\frac{y^2}{z+x}\ge\frac{y^2}{\sqrt{2\left(x^2+z^2\right)}},\frac{z^2}{x+y}\ge\frac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
\(\Rightarrow T\ge\frac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\frac{y^2}{\sqrt{2\left(x^2+z^2\right)}}+\frac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
Đặt : \(\sqrt{x^2+y^2}=a;\sqrt{y^2+z^2}=b;\sqrt{x^2+z^2}=c\left(a,b,c>0\right)\)
Khi đó: \(T\ge\frac{1}{2\sqrt{2}}\left(\frac{a^2+c^2-b^2}{b}+\frac{a^2+b^2-c^2}{c}+\frac{b^2+c^2-a^2}{a}\right)\)
\(\Leftrightarrow T\ge\frac{1}{2\sqrt{2}}\left(\left(\frac{\left(a+c\right)^2}{2b}-b\right)+\left(\frac{\left(a+b\right)^2}{2c}-c\right)+\left(\frac{\left(b+c\right)^2}{2a}-a\right)\right)\)
\(\ge\frac{1}{2\sqrt{2}}\left(2\left(a+c\right)-3b+2\left(a+b\right)-3c+2\left(b+c\right)-3a\right)\)
\(\Rightarrow T\ge\frac{1}{2\sqrt{2}}\left(a+b+c\right)=\frac{1}{2}\sqrt{\frac{2017}{2}}\)
de thay x,y co vi tri nhu nhau nen x=y suy ra p=0