trong đề thi HSG tỉnh thanh hóa năm 2010-2011(đánh lên mạng đi,hình như là bài 5)

Áp dụng BĐT \(\sqrt{a^2+b^2}\ge\frac{\sqrt{2}}{2}\left(a+b\right)\) (bạn tự chứng minh)

Ta có \(P=\frac{\sqrt{x^2+y^2}}{z}+\frac{\sqrt{y^2+z^2}}{x}+\frac{\sqrt{z^2+x^2}}{y}\ge\frac{\sqrt{2}}{2}\left(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\right)\)

\(=\frac{\sqrt{2}}{2}\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\right]\ge\frac{\sqrt{2}}{2}\left(2+2+2\right)=3\sqrt{2}\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}x=y=z\\x,y,z>0\end{cases}}\)

Vậy min P = \(3\sqrt{2}\) khi x = y = z

Nhớ mang máng câu này hồi trước có giải rồi. Thôi tự vô tìm đi nha

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:  \(T\ge\frac{1}{2\sqrt{2}}\left(\frac{a^2+c^2-b^2}{b}+\frac{a^2+b^2-c^2}{c}+\frac{b^2+c^2-a^2}{a}\right)\)

\(\Leftrightarrow T\ge\frac{1}{2\sqrt{2}}\left(\left(\frac{\left(a+c\right)^2}{2b}-b\right)+\left(\frac{\left(a+b\right)^2}{2c}-c\right)+\left(\frac{\left(b+c\right)^2}{2a}-a\right)\right)\)

\(\ge\frac{1}{2\sqrt{2}}\left(2\left(a+c\right)-3b+2\left(a+b\right)-3c+2\left(b+c\right)-3a\right)\)

\(\Rightarrow T\ge\frac{1}{2\sqrt{2}}\left(a+b+c\right)=\frac{1}{2}\sqrt{\frac{2017}{2}}\)

Đặt xong thì suy ra:

\(x^2=\frac{a^2+c^2-b^2}{2}\)

\(y^2=\frac{a^2+b^2-c^2}{2}\)

\(z^2=\frac{b^2+c^2-a^2}{2}\)

Phần sau thì thay vào rồi phân h ra thôi

\(\frac{\left(a+c\right)^2}{2b}-b=\frac{\left(a+c\right)^2}{2b}+2b-3b\ge2\left(a+c\right)-3b\left(bđtcô-si\right)\)

\(T\ge\frac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\frac{y^2}{\sqrt{2\left(x^2+z^2\right)}}+\frac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)

Đặt \(\left(\sqrt{y^2+z^2};\sqrt{x^2+z^2};\sqrt{x^2+y^2}\right)=\left(a;b;c\right)\Rightarrow a+b=c=2014\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\frac{b^2+c^2-a^2}{2}\\y^2=\frac{a^2+c^2-b^2}{2}\\z^2=\frac{a^2+b^2-c^2}{2}\end{matrix}\right.\)

\(\Rightarrow T.2\sqrt{2}\ge\frac{b^2+c^2-a^2}{a}+\frac{a^2+c^2-b^2}{b}+\frac{a^2+b^2-c^2}{c}\)

\(T.2\sqrt{2}\ge\frac{\left(b+c\right)^2}{2a}+\frac{\left(a+c\right)^2}{2b}+\frac{\left(a+b\right)^2}{2c}-\left(a+b+c\right)\)

\(T.2\sqrt{2}\ge\frac{4\left(a+b+c\right)^2}{2\left(a+b+c\right)}-\left(a+b+c\right)=a+b+c=2014\)

\(\Rightarrow T\ge\frac{1007}{\sqrt{2}}\)

Dấu "=" xảy ra khi \(x=y=z=...\)

\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

Đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=1\)

\(P=\dfrac{2a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}=\dfrac{2a}{\sqrt{ab+bc+ca+a^2}}+\dfrac{b}{\sqrt{ab+bc+ca+b^2}}+\dfrac{c}{\sqrt{ab+bc+ca+c^2}}\)

\(P=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(P=\sqrt{\dfrac{2a}{a+b}.\dfrac{2a}{a+c}}+\sqrt{\dfrac{2b}{a+b}.\dfrac{b}{2\left(b+c\right)}}+\sqrt{\dfrac{2c}{c+a}.\dfrac{c}{2\left(c+b\right)}}\)

\(P\le\dfrac{1}{2}\left(\dfrac{2a}{a+b}+\dfrac{2a}{a+c}+\dfrac{2b}{a+b}+\dfrac{b}{2\left(b+c\right)}+\dfrac{2c}{c+a}+\dfrac{c}{2\left(c+b\right)}\right)=\dfrac{9}{4}\)

\(P_{max}=\dfrac{9}{4}\) khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\) hay \(\left(x;y;z\right)=\left(\dfrac{\sqrt{15}}{7};\sqrt{15};\sqrt{15}\right)\)