Cho x,y,z > 0 và \(x+y+z\ge6\). Tìm GTNN của Q = \(\dfrac{x^3}{y+z}+\dfrac{y^3}{x+z}+\dfrac{z^3}{x+y}\)
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`P=x^3/(x+y)+y^3/(y+z)+z^3/(z+x)`
`=x^4/(x^2+xy)+y^4/(y^2+yz)+z^4/(z^2+zx)`
Ad bđt cosi-swart:
`P>=(x^2+y^2+z^2)^2/(x^2+y^2+z^2+xy+yz+zx)`
Mà `xy+yz+zx<=x^2+y^2+z^2)`
`=>P>=(x^2+y^2+z^2)^2/(2(x^2+y^2+z^2))=(x^2+y^2+z^2)/2=3/2`
Dấu "=" xảy ra khi `x=y=z=1`
`Q=(x^3+y^3)/(x+2y)+(y^3+z^3)/(y+2z)+(z^3+x^3)/(z+2x)`
`Q=(x^3/(x+2y)+y^3/(y+2z)+z^3/(z+2x))+(y^3/(x+2y)+z^3/(y+2z)+x^3/(z+2x))`
`Q=(x^4/(x^2+2xy)+y^4/(y^2+2yz)+z^4/(z^2+2zx))+(y^4/(xy+2y^2)+z^4/(yz+2z^4)+x^4/(xz+2x^2))`
Áp dụng BĐT cosi-swart ta có:
`Q>=(x^2+y^2+z^2)^2/(x^2+y^2+z^2+2xy+2yz+2zx)+(x^2+y^2+z^2)^2/(2(x^2+y^2+z^2)+xy+yz+zx))`
Mà`xy+yz+zx<=x^2+y^2+z^2`
`=>Q>=(x^2+y^2+z^2)^2/(3(x^2+y^2+z^2))+(x^2+y^2+z^2)^2/(3(x^2+y^2+z^2))=(2(x^2+y^2+z^2)^2)/(3(x^2+y^2+z^2))=(2(x^2+y^2+z^2))/3=2`
Dấu "=" xảy ra khi `x=y=z=1.`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\sqrt{\dfrac{x^3}{y+3}}+\sqrt{\dfrac{y^3}{z+3}}+\sqrt{\dfrac{z^3}{x+3}}\)
\(=\dfrac{x^2}{\sqrt{x\left(y+3\right)}}+\dfrac{y^2}{\sqrt{y\left(z+3\right)}}+\dfrac{z^2}{\sqrt{z\left(x+3\right)}}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}}\)
Xét:
\(\left(\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}\right)^2\le\left(1^2+1^2+1^2\right)\left(xy+3x+yz+3y+xz+3z\right)\)
\(=3\left(9+xy+yz+xz\right)\)
\(=27+3\left(xy+yz+xz\right)\le27+\left(x+y+z\right)^2=36\)
\(\Rightarrow\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}\le6\)
\(P\ge\dfrac{3}{2}\)
\("="\Leftrightarrow x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P+3=x+\left(y^2+1\right)+\left(z^3+1+1\right)\ge x+2y+3z\)
\(\Rightarrow P\ge x+2y+3z-3\)
\(6=\dfrac{1}{x}+\dfrac{4}{2y}+\dfrac{9}{3z}\ge\dfrac{\left(1+2+3\right)^2}{x+2y+3z}\)
\(\Rightarrow x+2y+3z\ge6\Rightarrow P\ge3\)
Dấu "=" xảy ra khi \(x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Áp dụng bđt Cauchy cho 3 số dương ta có
\(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)
\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)
\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)
Cộng (1);(2);(3) theo vế ta được
\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)
\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)
\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)
2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)
Dấu"=" khi a = 4b
nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)
Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)
Đặt \(\sqrt{a+b}=t>0\) ta được
\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)
\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)
Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)
nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)
khi đó a + b = 1
mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Bạn cần bổ sung điều kiện $x,y,z>0$
\(P=\frac{1}{x.\frac{y^2+z^2}{y^2z^2}}+\frac{1}{y.\frac{z^2+x^2}{z^2x^2}}+\frac{1}{z.\frac{x^2+y^2}{x^2y^2}}=\frac{1}{x(\frac{1}{y^2}+\frac{1}{z^2})}+\frac{1}{y(\frac{1}{z^2}+\frac{1}{x^2})}+\frac{1}{z(\frac{1}{x^2}+\frac{1}{y^2})}\)
\(=\frac{1}{x(3-\frac{1}{x^2})}+\frac{1}{y(3-\frac{1}{y^2})}+\frac{1}{z(3-\frac{1}{z^2})}=\frac{x}{3x^2-1}+\frac{y}{3y^2-1}+\frac{z}{3z^2-1}\)
Vì $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=3\Rightarrow x^2, y^2, z^2>\frac{1}{3}$
Xét hiệu:
\(\frac{x}{3x^2-1}-\frac{1}{2x^2}=\frac{(x-1)^2(2x+1)}{2x^2(3x^2-1)}\geq 0\) với mọi $x>0$ và $x^2>\frac{1}{3}$
$\Rightarrow \frac{x}{3x^2-1}\geq \frac{1}{2x^2}$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế ta có:
$P\geq \frac{1}{2}(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})=\frac{3}{2}$
Vậy $P_{\min}=\frac{3}{2}$ khi $x=y=z=1$
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Ta có: \(2x^3+2y^3-\left(x+y\right)\left(x^2+y^2\right)=\left(x-y\right)^2\left(x+y\right)\ge0\)
\(\Rightarrow\dfrac{x^3+y^3}{x^2+y^2}\ge\dfrac{x+y}{2}\)
Tương tự: \(\dfrac{y^3+z^3}{y^2+z^2}\ge\dfrac{y+z}{2}\) ; \(\dfrac{z^3+x^3}{z^2+x^2}\ge\dfrac{z+x}{2}\)
Cộng vế: \(P\ge x+y+z\ge6\)
\(P_{min}=6\) khi \(x=y=z=2\)
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\(x,y,z>0\)
Áp dụng BĐT Caushy cho 3 số ta có:
\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)
\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)
\(=\dfrac{\left(x^3-1\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)}+\dfrac{\left(y^3-1\right)^2}{\left(x+y^2+z\right)\left(y^3-1\right)}+\dfrac{\left(z^3-1\right)^2}{\left(x+y+z^2\right)\left(x^3-1\right)}\)
Áp dụng BĐT Caushy-Schwarz ta có:
\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)
\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)
\(P=0\Leftrightarrow x=y=z=1\)
Vậy \(P_{min}=0\)
\(Q=\dfrac{x^3}{y+z}+\dfrac{y^3}{x+z}+\dfrac{z^3}{x+y}\)
\(Q=\dfrac{x^4}{xy+xz}+\dfrac{y^4}{xy+zy}+\dfrac{z^4}{xz+yz}\)
\(Q\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+xz+xy+zy+xz+yz}=\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(xy+yz+xz\right)}\)(svac-xo)
Lại có:\(x^2+y^2+z^2\ge xy+yz+zx\)(tự cm)
\(\Rightarrow Q\ge\dfrac{x^2+y^2+z^2}{2}\)
Mặt khác:\(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\ge36\)(tự cm)
\(\Rightarrow x^2+y^2+z^2\ge12\)
\(\Rightarrow Q\ge\dfrac{12}{2}=6\)
Vậy MINQ=6<=>x=y=z=2
Ta có: \((\dfrac{x^3}{y+z}+\dfrac{y+z}{x})+\left(\dfrac{y^3}{x+z}+\dfrac{x+z}{y}\right)+\left(\dfrac{z^3}{x+y}+\dfrac{x+y}{z}\right)\ge2\sqrt{\dfrac{x^3\left(y+z\right)}{\left(y+z\right)x}}+2\sqrt{\dfrac{y^3\left(x+z\right)}{\left(x+z\right)y}}+2\sqrt{\dfrac{z^3\left(x+y\right)}{\left(x+y\right)z}}=2\sqrt{x^2}+2\sqrt{y^2}+2\sqrt{z^2}=2\left(x+y+z\right)\ge2.6=12\)
(Bất đẳng thức cauchy)
mà \(\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}=\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{x}{y}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{y}{z} \)
\(=\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge2\sqrt{\dfrac{yx}{xy}}+2\sqrt{\dfrac{zx}{xz}}+2\sqrt{\dfrac{zy}{yz}}=2+2+2=6\) (Bất đẳng thức cauchy)
\(\Rightarrow P\ge12-6=6\)
Dấu "=" xảy ra \(\Leftrightarrow\)x = y = z = 2
Vậy GTNN của P = 6 \(\Leftrightarrow\)x = y = z = 2