\(P=\dfrac{x}{\sqrt{x+y-x}}+\dfrac{y}{\sqrt{x+y-y}}=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\)

\(=\dfrac{x^2}{x\sqrt{y}}+\dfrac{y^2}{y\sqrt{x}}\ge\dfrac{\left(x+y\right)^2}{x\sqrt{y}+y\sqrt{x}}=\dfrac{\left(x+y\right)^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)

\(\ge\dfrac{\left(x+y\right)^2}{\dfrac{x+y}{2}.\left(1.\sqrt{x}+1.\sqrt{y}\right)}\ge\dfrac{\left(x+y\right)^2}{\dfrac{x+y}{2}.\sqrt{\left(1^2+1^2\right)\left(x+y\right)}}=\dfrac{1}{\dfrac{1}{2}\sqrt{2}}=\sqrt{2}\)

"=" khi x = y = 1/2

giúp mình voi ah

\(T=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}=\dfrac{x^2}{x\sqrt{y}}+\dfrac{y^2}{y\sqrt{x}}\ge\dfrac{\left(x+y\right)^2}{x\sqrt{y}+y\sqrt{x}}=\dfrac{1}{x\sqrt{y}+y\sqrt{x}}\)

\(\Rightarrow T\ge\dfrac{1}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\ge\dfrac{1}{\dfrac{\left(x+y\right)}{2}.\sqrt{2\left(x+y\right)}}=\sqrt{2}\)

\(\Rightarrow T_{min}=\sqrt{2}\) khi \(x=y=\dfrac{1}{2}\)

Giúp mình câu này với ah. 

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\geq \frac{(x+y+z)^2}{x+y+y+z+z+x}\)

\(\Leftrightarrow A\geq \frac{x+y+z}{2}\)

Áp dụng BĐT AM-GM:

\(\left\{\begin{matrix} x+y\geq 2\sqrt{xy}\\ y+z\geq 2\sqrt{yz}\\ z+x\geq 2\sqrt{zx}\end{matrix}\right.\)

\(\Rightarrow 2(x+y+z)\geq 2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})=2\)

\(\Rightarrow x+y+z\geq 1\)

Do đó: \(A\geq \frac{x+y+z}{2}\geq \frac{1}{2}\)

Vậy \(A_{\min}=\frac{1}{2}\)

Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)

b, đk: \(x\ge1,y\ge2,z\ge3\)

\(=>B=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)

đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{y-2}=b\\\sqrt{z-3}=c\end{matrix}\right.\)\(=>\left\{{}\begin{matrix}x=a^2+1\\y=b^2+1\\z=c^2+1\end{matrix}\right.\)\(=>a\ge0,b\ge0,c\ge0\)

B trở thành \(\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\)

\(=\dfrac{a^{ }}{a^2+1}+\dfrac{a^2+1}{4}+\dfrac{b}{b^2+1}+\dfrac{b^2+1}{4}+\dfrac{c}{c^2+1}+\dfrac{c^2+1}{4}\)

\(-\left(\dfrac{a^2+b^2+c^2+3}{4}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}-\dfrac{a^2+b^2+c^2}{4}\)\(=0\)

dấu"=" xảy ra<=>\(a=0,b=0,c=0< =>x=1,y=2,z=3\)

Chắc bạn ghi nhầm đề, tìm GTLN mới đúng, chứ GTNN của các biểu thức này đều hiển nhiên bằng 0

\(A=\dfrac{3.\sqrt{x-9}}{15x}\le\dfrac{3^2+x-9}{30x}=\dfrac{1}{30}\)

\(A_{max}=\dfrac{1}{30}\) khi \(x=18\)

\(B=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}=\dfrac{1.\sqrt{x-1}}{x}+\dfrac{\sqrt{2}.\sqrt{y-2}}{\sqrt{2}y}+\dfrac{\sqrt{3}.\sqrt{z-3}}{\sqrt{3}z}\)

\(B\le\dfrac{1+x-1}{2x}+\dfrac{2+y-2}{2\sqrt{2}y}+\dfrac{3+z-3}{2\sqrt{3}z}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;4;6\right)\)

đề bài là tìm gt lớn nhất nhé mọi người,tớ ghi nhầm

ezezezezezezezez

              Bài làm :

ĐKXĐ : \(x-2\ge0\Rightarrow x\ge2\)

Ta có ;

\(P=...\)

\(=x^2-2xy+y^2+x^2-4x+4+x+\frac{1}{x}+2\sqrt{x-2}+2017\)

\(=\left(x-y\right)^2+\left(x-2\right)^2+\frac{x}{4}+\frac{1}{x}+\frac{3x}{4}+2\sqrt{x-2}+2017\)

Áp dụng BĐT cosi cho các số không âm ; ta có :

\(x+\frac{1}{x}=\frac{x}{4}+\frac{1}{x}+\frac{3}{4}x\ge2\sqrt{\frac{x}{4}.\frac{1}{x}}+\frac{3}{4}.2=\frac{5}{2}\)

\(\Rightarrow P\ge2017+\frac{5}{2}=\frac{4039}{2}\)

Dấu "=" xảy ra khi : x=y=2

Vậy Min (P) = 4039/2 khi : x=y=2

Ta có: \(P=\sqrt{a^2+a}+\sqrt{b^2+b}+\sqrt{c^2+c}\)

\(=\sqrt{a\left(a+1\right)}+\sqrt{b\left(b+1\right)}+\sqrt{c\left(c+1\right)}\)

\(=\frac{1}{2}\left[\sqrt{4a\left(a+1\right)}+\sqrt{4b\left(b+1\right)}+\sqrt{4c\left(c+1\right)}\right]\)

\(\le\frac{1}{2}\left(\frac{4a+a+1}{4}+\frac{4b+b+1}{4}+\frac{4c+c+1}{4}\right)\)

\(=\frac{1}{2}\cdot\frac{5\left(a+b+c\right)+3}{4}=\frac{1}{2}\cdot4=2\)

Dấu "=" xảy ra khi: a = b = c = 1/3

Lại có: \(0\le a,b,c\le1\Rightarrow\hept{\begin{cases}a\ge a^2\\b\ge b^2\\c\ge c^2\end{cases}}\)

\(\Rightarrow P\ge\sqrt{a^2+a^2}+\sqrt{b^2+b^2}+\sqrt{c^2+c^2}=\sqrt{2}\left(a+b+c\right)=\sqrt{2}\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}a=1\\b=c=0\end{cases}}\) và các hoán vị