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![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi cái biểu thức đó là P nha
Trước tiên chứng minh:
\(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\left(\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\right)=0\)
\(\Leftrightarrow\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4-z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4-x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(\Leftrightarrow x-y+y-z+z-x=0\)( đúng )
Giờ ta quay lại bài toán ban đầu
Ta có:
\(\Leftrightarrow2P=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(\ge\frac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}+\frac{\left(y^2+z^2\right)^2}{2\left(y^2+z^2\right)\left(y+z\right)}+\frac{\left(z^2+x^2\right)^2}{2\left(z^2+x^2\right)\left(z+x\right)}\)
\(=\frac{x^2+y^2}{2\left(x+y\right)}+\frac{y^2+z^2}{2\left(y+z\right)}+\frac{z^2+x^2}{2\left(z+x\right)}\)
\(\ge\frac{\left(x+y\right)^2}{4\left(x+y\right)}+\frac{\left(y+z\right)^2}{4\left(y+z\right)}+\frac{\left(z+x\right)^2}{4\left(z+x\right)}\)
\(=\frac{x+y}{4}+\frac{y+z}{4}+\frac{z+x}{4}=\frac{1}{2}\)
\(\Rightarrow P\ge\frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Xét hiệu:
\(\frac{x^4}{(x^2+y^2)(x+y)}+\frac{y^4}{(y^2+z^2)(y+z)}+\frac{z^4}{(z^2+x^2)(z+x)}-\left(\frac{y^4}{(x^2+y^2)(x+y)}+\frac{z^4}{(y^2+z^2)(y+z)}+\frac{x^4}{(z^2+x^2)(z+x)}\right)\)
\(=\frac{x^4-y^4}{(x^2+y^2)(x+y)}+\frac{y^4-z^4}{(y^2+z^2)(y+z)}+\frac{z^4-x^4}{(z^2+x^2)(z+x)}\)
\(=x-y+y-z+z-x=0\)
\(\Rightarrow \frac{x^4}{(x^2+y^2)(x+y)}+\frac{y^4}{(y^2+z^2)(y+z)}+\frac{z^4}{(z^2+x^2)(z+x)}=\frac{y^4}{(x^2+y^2)(x+y)}+\frac{z^4}{(y^2+z^2)(y+z)}+\frac{x^4}{(z^2+x^2)(z+x)}\)
Do đó:
\(2F=\frac{x^4+y^4}{(x^2+y^2)(x+y)}+\frac{y^4+z^4}{(y^2+z^2)(y+z)}+\frac{z^4+x^4}{(z^2+x^2)(z+x)}\)
\(\geq \frac{\frac{(x^2+y^2)^2}{2}}{(x^2+y^2)(x+y)}+\frac{\frac{(y^2+z^2)^2}{2}}{(y^2+z^2)(y+z)}+\frac{\frac{(z^2+x^2)^2}{2}}{(z^2+x^2)(z+x)}\) (áp dụng BĐT Cauchy)
hay \(2F\geq \frac{x^2+y^2}{2(x+y)}+\frac{y^2+z^2}{2(y+z)}+\frac{z^2+x^2}{2(z+x)}\)
Mà cũng theo BĐT Cauchy thì:
\(\frac{x^2+y^2}{2(x+y)}+\frac{y^2+z^2}{2(y+z)}+\frac{z^2+x^2}{2(z+x)}\geq \frac{\frac{(x+y)^2}{2}}{2(x+y)}+\frac{\frac{(y+z)^2}{2}}{2(y+z)}+\frac{\frac{(z+x)^2}{2}}{2(x+z)}=\frac{x+y+z}{2}=\frac{1}{2}\)
\(\Rightarrow 2F\geq \frac{1}{2}\Rightarrow F\geq \frac{1}{4}\)
Vậy \(F_{\min}=\frac{1}{4}\Leftrightarrow x=y=z=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét: \(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}-\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}=\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}\)\(=\frac{\left(x^2+y^2\right)\left(x^2-y^2\right)}{\left(x^2+y^2\right)\left(x+y\right)}=\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(x^2+y^2\right)\left(x+y\right)}=x-y\)(1)
Tương tự, ta có: \(\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}-\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}=y-z\)(2); \(\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}=z-x\)(3)
Cộng theo vế của 3 đẳng thức (1), (2), (3), ta được:
\(\left[\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\right]\)\(-\left[\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\right]=0\)
\(\Rightarrow\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\)\(=\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
Mà \(A=\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\)nên \(2A=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)\(\ge\frac{\frac{\left(y^2+z^2\right)^2}{2}}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{\frac{\left(y^2+z^2\right)^2}{2}}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{\frac{\left(z^2+x^2\right)^2}{2}}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(=\frac{1}{2}\left(\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}\right)\)\(\ge\frac{1}{2}\left(\frac{\frac{\left(x+y\right)^2}{2}}{x+y}+\frac{\frac{\left(y+z\right)^2}{2}}{y+z}+\frac{\frac{\left(z+x\right)^2}{2}}{z+x}\right)\)\(=\frac{1}{4}\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]=\frac{1}{2}\left(x+y+z\right)=\frac{1}{2}\)(Do theo giả thiết thì x + y + z = 1)
\(\Rightarrow A\ge\frac{1}{4}\)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)
Bài này t làm rồi, "nhẹ" không ấy mà :|
Dự đoán khi \(x=y=z=\frac{1}{3}\Rightarrow A=\frac{1}{4}\). Ta c/m nó là GTNN của A
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(A=Σ\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{Σ\left(x^2+y^2\right)\left(x+y\right)}\)
Cần chứng minh BĐT \(\frac{\left(x^2+y^2+z^2\right)^2}{Σ\left(x^2+y^2\right)\left(x+y\right)}\ge\frac{x+y+z}{4}\)
\(\Leftrightarrow4\left(x^2+y^2+z^2\right)^2\ge\left(x+y+z\right)Σ\left(2x^3+x^2y+x^2z\right)\)
\(\LeftrightarrowΣ\left(2x^4-3x^3y-3x^3z+6x^2y^2-2x^2yz\right)\ge0\)
\(\LeftrightarrowΣ\left(2x^4-3x^3y-3x^3z+4x^2y^2\right)+Σ\left(2x^2y^2-2x^2yz\right)\ge0\)
\(\LeftrightarrowΣ\left(x^4-3x^3y+4x^2y^2-3xy^3+y^4\right)+Σ\left(x^2z^2-2z^2xy+y^2z^2\right)\ge0\)
\(\LeftrightarrowΣ\left(x-y\right)^2\left(x^2-xy+y^2\right)+Σz^2\left(x-y\right)^2\ge0\)
BĐT cuối đúng tức ta có \(A_{Min}=\frac{1}{4}\Leftrightarrow x=y=z=\frac{1}{3}\)
P/s: Nguồn lời giải Câu hỏi của Vo Trong Duy - Toán lớp 9 - Học toán với OnlineMath, rảnh quá ngồi gõ lại :V
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)
\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)
\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Hướng dẫn: đặt \(A=\dfrac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
Khi đó \(F-A=x-y+y-z+z-x=0\Rightarrow F=A\)
\(\Rightarrow2F=F+A=\sum\dfrac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x+y\right)^2\left(x^2+y^2\right)}{4\left(x^2+y^2\right)\left(x+y\right)}\)
\(\Rightarrow2F\ge\dfrac{x+y+z}{2}\Rightarrow F\ge\dfrac{x+y+z}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo đề bài ta có:
\(2\left(y^2+1\right)+6\ge\left(x^4+1\right)+\left(y^4+4\right)+\left(z^4+1\right)\ge2x^2+4y^2+2z^2\)
\(\Rightarrow0< x^2+y^2+z^2\le4\)
Đặt: \(t=x^2+y^2+z^2.Đkxđ:0< t\le4\)
Ta có: \(\sqrt{2}\left(x+y\right)y=\sqrt{2x}y+\sqrt{2z}y\le\frac{2x^2+y^2}{2}+\frac{2z^2+y^2}{2}=x^2+y^2+z^2\)
\(P\le x^2+y^2+z^2+\frac{1}{x^2+y^2+z^2+1}=t+\frac{1}{t+1}=f\left(t\right)\)
Xét hàm: \(f\left(t\right)=t+\frac{1}{t+1}\) liên tục trên \(\left(0;4\right)\)
\(f'\left(t\right)=1-\frac{1}{\left(t+1\right)^2}>0\forall t\in\left\{0;4\right\}\)nên:
\(\Rightarrow f\left(t\right)\) đồng biến trên \(\left\{0;4\right\}\)
\(\Rightarrow P\le f\left(t\right)\le f\left(4\right)=\frac{21}{5}\forall t\in\left(0;4\right)\)
\(\Rightarrow P_{Min}=\frac{21}{5}\Leftrightarrow\orbr{\begin{cases}x=z=1\\y=\sqrt{2}\end{cases}}\)
Vậy ....................
ミ★๖ۣۜBăηɠ ๖ۣۜBăηɠ ★彡
có cách nào không dùng hàm k ???
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(x+z\right)}\)
\(\Rightarrow F-A=\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^2-z^2}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4-x^4}{\left(z^2+x^2\right)\left(z+x\right)}=0\)
\(\Rightarrow F=A\)
\(\Rightarrow2F=F+A=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(\Rightarrow2F\ge\frac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}+\frac{\left(y^2+z^2\right)^2}{2\left(y^2+z^2\right)\left(y+z\right)}+\frac{\left(z^2+x^2\right)^2}{2\left(z^2+x^2\right)\left(z+x\right)}\)
\(\Rightarrow2F\ge\frac{x^2+y^2}{2\left(x+y\right)}+\frac{y^2+z^2}{2\left(y+z\right)}+\frac{z^2+x^2}{2\left(z+x\right)}\ge\frac{\left(x+y\right)^2}{4\left(x+y\right)}+\frac{\left(y+z\right)^2}{4\left(y+z\right)}+\frac{\left(z+x\right)^2}{4\left(z+x\right)}\)
\(\Rightarrow2F\ge\frac{1}{2}\left(x+y+z\right)=\frac{1}{2}\Rightarrow F\ge\frac{1}{4}\)
\(F_{min}=\frac{1}{4}\) khi \(x=y=z=\frac{1}{3}\)
Ta có : \(P=x^3+x^2y+y^3+y^2z+z^3+z^2x\)
\(=x^3+y^3+z^3+x^2y+y^2z+z^2x\)
Áp dụng BĐT Cô-si cho 3 số, ta có : \(x^2y=x.x.y\le\frac{x^3+y^3+z^3}{3}\)
tương tự : \(y^2z\le\frac{y^3+y^3+z^3}{3}\); \(z^2x\le\frac{z^3+z^3+x^3}{3}\)
\(\Rightarrow x^2y+y^2z+z^2x\le\frac{3\left(x^3+y^3+z^3\right)}{3}=x^3+y^3+z^3\)
\(\Rightarrow P\le2\left(x^3+y^3+z^3\right)\)
Áp dụng BĐT Cô-si cho 4 số, ta có : \(x^4+x^4+x^4+1\ge4\sqrt[4]{\left(x^4\right)^3.1}=4x^3\)
\(\Rightarrow3x^4+1\ge4x^3\)
Tương tự : \(3y^4+1\ge4y^3;3z^4+1\ge4z^3\)
Cộng lại theo vế, ta được : \(3\left(x^4+y^4+z^4\right)+3\ge4\left(x^3+y^3+z^3\right)\)
\(\Rightarrow2P\le4\left(x^3+y^3+z^3\right)\le3\left(x^4+y^4+z^4\right)+3=12\)
\(\Rightarrow P\le6\)
Vậy GTLN của P là 6 khi x = y = z = 1
Giả sử \(y=min\left\{x,y,z\right\}\)
Đẳng thức xảy ra khi \(x=y=z=1\)