Thử nhé

Vì P là bất đẳng thức đối xứng nên dự đoán điểm rơi \(x=y=z=\dfrac{\sqrt{2021}}{3}\)

Thay vo P ta duoc \(P=4.\sqrt{2021}\)

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\(P=\sum\dfrac{\left(x+y\right)\sqrt{\left(y+z\right)\left(z+x\right)}}{z}\)

Cauchy-Schwarz:

\(\Rightarrow\left(y+z\right)\left(z+x\right)\ge\left(z+\sqrt{xy}\right)^2\Rightarrow\sqrt{\left(y+z\right)\left(z+x\right)}\ge z+\sqrt{xy}\)

\(\Rightarrow P\ge\sum\dfrac{\left(x+y\right)\left(z+\sqrt{xy}\right)}{z}\ge\sum\dfrac{xz+yz+x\sqrt{y}+y\sqrt{x}}{z}=\sum x+y+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge\sum x+y+\dfrac{2xy}{z}\)

\(\Rightarrow P\ge2(x+y+z)+2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\)

Cauchy-Schwarz: \(\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\ge\left(\sqrt{\dfrac{xy}{z}.\dfrac{yz}{z}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)^2=\left(x+y+z\right)^2\)

\(\Rightarrow P\ge2(x+y+z)+2\left(x+y+z\right)=4\left(x+y+z\right)=4\sqrt{2021}\)

\("="\Leftrightarrow x=y=z=\dfrac{\sqrt{2021}}{3}\)

bạn vào trang này nhé có bài như thến này đấy 

//123doc.org//document/3173507-ren-luyen-chuyen-de-tim-maxmin-on-thi-thpt-quoc-gia.htm

tính diện tích hình vẽ dưới đây

Áp dụng bất đẳng thức cauchy:

\(P=\sum\dfrac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}\ge\sum\dfrac{2x^2\sqrt{yz}}{y\sqrt{y}+2z\sqrt{z}}=\sum\dfrac{2\sqrt{x^3}\sqrt{xyz}}{\sqrt{y^3}+2\sqrt{z^3}}=\sum\dfrac{2\sqrt{x^3}}{\sqrt{y^3}+2\sqrt{z^3}}\)(vì xyz=1).

đặt \(\left\{{}\begin{matrix}\sqrt{x^3}=a\\\sqrt{y^3}=b\\\sqrt{z^3}=c\end{matrix}\right.\)(\(a,b,c>0\))thì giả thiết trở thành cho abc=1. tìm Min \(P=\dfrac{2a}{b+2c}+\dfrac{2b}{c+2a}+\dfrac{2c}{a+2b}\)

Áp dụng BĐT cauchy-schwarz:

\(P=2\left(\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ac+2bc}\right)\ge\dfrac{2\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=2\)( AM-GM \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\))

Dấu = xảy ra khi a=b=c=1 hay x=y=z=1

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((x+y)(x+z)\geq (x+\sqrt{yz})^2\)

\(\Rightarrow \sqrt{(x+y)(y+z)(x+z)}.\frac{\sqrt{y+z}}{x}\geq \frac{(y+z)(x+\sqrt{yz})}{x}=y+z+\frac{\sqrt{yz}(y+z)}{x}\)

Hoàn toàn tương tự :

\(\sqrt{(x+y)(y+z)(x+z)}.\frac{\sqrt{x+z}}{y}\geq x+z+\frac{\sqrt{xz}(x+z)}{y}\)

\(\sqrt{(x+y)(y+z)(x+z)}.\frac{\sqrt{x+y}}{z}\geq x+y+\frac{\sqrt{xy}(x+y)}{z}\)

Cộng theo vế:

\(T\geq 2(x+y+z)+\underbrace{\frac{(x+y)\sqrt{xy}}{z}+\frac{(y+z)\sqrt{yz}}{x}+\frac{(z+x)\sqrt{zx}}{y}}_{M}\)

Ta có:

\(M=\frac{(\sqrt{2}-z)\sqrt{xy}}{z}+\frac{(\sqrt{2}-x)\sqrt{yz}}{x}+\frac{(\sqrt{2}-y)\sqrt{xz}}{y}\)

\(=\sqrt{2}\left(\frac{\sqrt{xy}}{z}+\frac{\sqrt{yz}}{x}+\frac{\sqrt{xz}}{y}\right)-(\sqrt{xy}+\sqrt{yz}+\sqrt{xz})\)

Áp dụng BĐT AM-GM:

\(\frac{\sqrt{xy}}{z}+\frac{\sqrt{yz}}{x}+\frac{\sqrt{xz}}{y}\geq 3\sqrt[3]{\frac{xyz}{xyz}}=3\)

\(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\leq \frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=x+y+z=\sqrt{2}\)

Do đó: \(M\geq 3\sqrt{2}-\sqrt{2}=2\sqrt{2}\)

\(\Rightarrow T\geq 2(x+y+z)+M\geq 2\sqrt{2}+2\sqrt{2}=4\sqrt{2}\)

Vậy \(T_{\min}=4\sqrt{2}\)

SEIFWJNHGRHFQ24FTW

Ta có: \(\sqrt{\left(x^2+\dfrac{1}{y^2}\right)\left(1+81\right)}\ge\sqrt{\left(x+\dfrac{9}{y}\right)^2}\)

=> \(\sqrt{x^2+\dfrac{1}{y^2}}\ge\dfrac{x+\dfrac{9}{y}}{\sqrt{82}}\)

Tương tự => \(\left\{{}\begin{matrix}\sqrt{y^2+\dfrac{1}{z^2}}\ge\dfrac{y+\dfrac{9}{z}}{\sqrt{82}}\\\sqrt{z^2+\dfrac{1}{x^2}}\ge\dfrac{z+\dfrac{9}{x}}{\sqrt{82}}\end{matrix}\right.\)

=> \(P\ge\dfrac{\left(x+y+z\right)+9\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}{\sqrt{82}}\)

Mà x + y + z = 1

      \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=9\)

=> \(P\ge\sqrt{82}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)