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Giải
Dự đoán dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\) khi đó \(P=\frac{3\sqrt{3}}{4}\)
Ta sẽ chứng minh nó là GTNN của \(P\)
Ta có: \(x^2+xy+y^2=\frac{3\left(x+y\right)^2+\left(x-y\right)^2}{4}\ge\frac{3\left(x+y\right)^2}{4}\)
Do đó ta cần chứng minh
\(\frac{x+y}{4yz+1}+\frac{y+z}{4xz+1}+\frac{x+z}{4xy+1}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{x+y}{\left(y+z\right)^2+1}+\frac{y+z}{\left(x+z\right)^2+1}+\frac{x+z}{\left(x+y\right)^2+1}\ge\frac{3}{2}\)
Ta có: \(x+y+z=\frac{3}{2}\Rightarrow2x+2y+2z=3\)
\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(x+z\right)=2\)
Đặt \(\hept{\begin{cases}a=x+y\\b=y+z\\c=z+x\end{cases}}\) thì ta cần chứng minh
\(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\ge\frac{3}{2}\)\(\forall\hept{\begin{cases}a,b,c>0\\a+b+c=3\end{cases}}\)
Lại có: \(\frac{a}{b^2+1}=a-\frac{ab^2}{b^2+1}\ge a-\frac{ab}{2}\)
Tương tự ta cũng có: \(\frac{b}{c^2+1}\ge b-\frac{bc}{2};\frac{c}{a^2+1}\ge c-\frac{ac}{2}\)
Cộng theo vế các BĐT ta có: \(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\ge a-\frac{ab}{2}+b-\frac{bc}{2}+c-\frac{ac}{2}\)
\(=\left(a+b+c\right)-\frac{ab+bc+ca}{2}\ge3-\frac{3}{2}=\frac{3}{2}\)
BĐT đã được c/m vậy ta có \(P\ge\frac{3\sqrt{3}}{4}\Leftrightarrow x=y=z=\frac{1}{2}\)
Bạn tham khảo tại đây:
Câu hỏi của Vũ Sơn Tùng - Toán lớp 9 | Học trực tuyến
\(M^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2xy}{\sqrt{yz}}+\frac{2yz}{\sqrt{zx}}+\frac{2xz}{\sqrt{yz}}=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2x\sqrt{y}}{\sqrt{z}}+\frac{2y\sqrt{z}}{\sqrt{x}}+\frac{2z\sqrt{x}}{\sqrt{y}}\)
Áp dụng bđt Cô-si: \(\frac{x^2}{y}+\frac{x\sqrt{y}}{\sqrt{z}}+\frac{x\sqrt{y}}{\sqrt{z}}+z\ge4\sqrt[4]{\frac{x^2}{y}.\frac{x\sqrt{y}}{\sqrt{z}}.\frac{x\sqrt{y}}{\sqrt{z}}.z}=4x\)
tương tự \(\frac{y^2}{z}+\frac{y\sqrt{z}}{\sqrt{x}}+\frac{y\sqrt{z}}{\sqrt{x}}+x\ge4y\);\(\frac{z^2}{x}+\frac{z\sqrt{x}}{\sqrt{y}}+\frac{z\sqrt{x}}{\sqrt{y}}+y\ge4z\)
=>\(M^2+x+y+z\ge4\left(x+y+z\right)\Rightarrow M^2\ge3\left(x+y+z\right)\ge3.12=36\Rightarrow M\ge6\)
Dấu "=" xảy ra khi x=y=z=4
Vậy minM=6 khi x=y=z=4
Lời giải:
Ta có:
\(x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\)
Hoàn toàn tương tự:
\(y^2+1=(y+z)(y+x); z^2+1=(z+x)(z+y)\)
Do đó:
\(\text{VT}=\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}=\frac{1}{(x+y)(x+z)}+\frac{1}{(y+z)(y+x)}+\frac{1}{(z+x)(z+y)}=\frac{2(x+y+z)}{(x+y)(y+z)(x+z)}(*)\)
----------------------------------------------------
\(\text{VP}=\frac{2}{3}\left(\frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\right)^3=\frac{2}{3}\left(\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(y+x)(y+z)}}+\frac{z}{\sqrt{(z+x)(z+y)}}\right)^3\)
\(=\frac{2}{3}.\frac{(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y})^3}{\sqrt{(x+y)(y+z)(x+z)}^3}(1)\)
Áp dụng BĐT Bunhiacopxky:
\((x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y})^2\leq (x+y+z)(xy+xz+yx+yz+zx+zy)=2(x+y+z)\)
\(\Rightarrow (x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y})^3\leq \sqrt{2(x+y+z)}^3(2)\)
\((x+y)(y+z)(x+z)=(x+y+z)(xy+yz+xz)-xyz\geq (x+y+z)(xy+yz+xz)-\frac{(x+y+z)(xy+yz+xz)}{9}\) (AM-GM)
\(=\frac{8}{9}(x+y+z)(xy+yz+xz)=\frac{8}{9}(x+y+z)\)
\(\Rightarrow \sqrt{(x+y)(y+z)(x+z)}^3\geq (x+y)(y+z)(x+z)\sqrt{\frac{8}{9}(x+y+z)}(3)\)
Từ \((1);(2);(3)\Rightarrow \text{VP}\leq \frac{2}{3}.\frac{\sqrt{2(x+y+z)}^3}{(x+y)(y+z)(x+z)\sqrt{\frac{8}{9}(x+y+z)}}=\frac{2(x+y+z)}{(x+y)(y+z)(x+z)}(**)\)
Từ \((*); (**)\Rightarrow \text{VT}\geq \text{VP}\). Ta có đpcm.
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
Ta có:
\(1+x^2=xy+yz+xz+x^2=\left(x+y\right)\left(x+z\right)\)
\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)
\(1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào T ta được:
\(T=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(=xy+xz+xy+yz+xz+zy\)
\(=2\left(xy+yz+xz\right)=2\left(xy+yz+xz=1\right)\)
Ta có \(1+x^2=x^2+xy+yz+zx=\left(x+y\right)\left(z+x\right)\).
Tương tự ta cũng có \(1+y^2=\left(x+y\right)\left(y+z\right)\) và \(1+z^2=\left(z+x\right)\left(y+z\right)\).
Thu gọn được \(T=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)
\(x^2+xy+y^2=\left(x+y\right)^2-xy\ge\left(x+y\right)^2-\frac{1}{4}\left(x+y\right)^2=\frac{3}{4}\left(x+y\right)^2\)
\(\Rightarrow\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}{2}\left(x+y\right)\)
Vậy:
\(P\ge\frac{\sqrt{3}}{2}\left[\frac{\left(x+y\right)^2}{1+4xy}+\frac{\left(y+z\right)^2}{1+4yz}+\frac{\left(z+x\right)^2}{1+4zx}\right]\)
\(P\ge\frac{\sqrt{3}}{2}\left[\frac{\left(2x+2y+2z\right)^2}{3+4\left(xy+yz+zx\right)}\right]\ge\frac{\sqrt{3}}{2}.\frac{9}{3+\frac{4}{3}\left(x+y+z\right)^2}=\frac{3\sqrt{3}}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)
\(\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}{2}\left(x+y\right)\) mà sao thế vào là \(\frac{\sqrt{3}}{2}\left(x+y\right)^2\) vậy ạ?