Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
`<=>2P=10x+6y+24/x+32/y`
`<=>2P=6x+24/x+2y+32/y+4x+4y`
`<=>2P=6(x+4/x)+2(y+16/y)+4(x+y)`
Áp dụng BĐT cosi:
`x+4/x>=4=>6(x+4/x)>=24`
`y+16/y>=8=>2(y+16/y)>=16`
Mà `x+y>=6=>4(x+y)>=24`
`=>2P>=24+16+24=64`
`=>P>=32`
Dấu "=" `<=>x=2,y=4`
P=\(5x+3y+\dfrac{12}{x}+\dfrac{16}{y}\)
=\(3x+\dfrac{12}{x}+y+\dfrac{16}{y}+2\left(x+y\right)\)
AD BĐT cô si :
Ta có \(3x+\dfrac{12}{x}\ge2\sqrt{3x.\dfrac{12}{x}}=2\sqrt{36}=12\)
\(y+\dfrac{16}{y}\ge2\sqrt{y.\dfrac{16}{y}}=2\sqrt{16}=8\)
\(2\left(x+y\right)\ge2.6=12\)
=> P\(\ge12+8+12=32\)
Dấu = xra \(\left\{{}\begin{matrix}3x=\dfrac{12}{x}\\y=\dfrac{16}{y}\\x+y=6\end{matrix}\right.\)\(\Leftrightarrow\left(x;y\right)=\left(2;4\right)\)
Vậy GTNN của P=32 khi (x;y)=(2;4)
\(S=\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y^3}{16\left(x+16\right)}+\dfrac{2021}{2022}\)
\(\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y+16}{100}+\dfrac{16}{80}\ge3\sqrt[3]{\dfrac{x^3\left(y+16\right).16}{16\left(y+16\right).100.80}}=\dfrac{3x}{20}\)
\(tương\) \(tự\Rightarrow\dfrac{y^3}{16\left(x+16\right)}\ge\dfrac{3y}{20}\)
\(\Rightarrow S\ge\dfrac{3x}{20}+\dfrac{3y}{20}-\left(\dfrac{x+16}{100}+\dfrac{y+16}{100}\right)-2.\dfrac{16}{80}+\dfrac{2021}{2022}=\dfrac{3x+3y}{20}-\dfrac{x+y+32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{15x+15y-x-y-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{14\left(x+y\right)-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}\)
\(xy=16\le\dfrac{\left(x+y\right)^2}{4}\Rightarrow x+y\ge8\Rightarrow S\ge\dfrac{14.8-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{2}{5}+\dfrac{2021}{2022}\)
\(\Rightarrow minS=\dfrac{2}{5}+\dfrac{2021}{2022}\Leftrightarrow x=y=4\)
\(\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y+16}{100}+\dfrac{1}{5}\ge3\sqrt[3]{\dfrac{x^3\left(y+16\right)}{16.100.5\left(y+16\right)}}=\dfrac{3x}{20}\)
Tương tự: \(\dfrac{y^3}{16\left(x+16\right)}+\dfrac{x+16}{100}+\dfrac{1}{5}\ge\dfrac{3y}{20}\)
Cộng vế:
\(S+\dfrac{x+y+32}{100}+\dfrac{2}{5}\ge\dfrac{3\left(x+y\right)}{20}+\dfrac{2021}{2022}\)
\(S\ge\dfrac{9}{20}\left(x+y\right)-\dfrac{42}{25}+\dfrac{2021}{2022}\ge\dfrac{9}{20}.2\sqrt{xy}-\dfrac{42}{25}+\dfrac{2021}{2022}=...\)
P=5x+3y+12/x+16/y
=3x+12/x+y+16/y+2(x+y)
áp dụng cosi: 3x+12/x>=2√(3.12)=12
y+16/y>=8
lại có 2(x+y)>=2.6=12
nên
P>=12+8+12=32
dấu = khi 3x=12/x và y=16/y và x+y=6
==> x=2; y=4
giá trị nhỏ nhất P=32 khi x=2; y=4
@Nguồn: Yahoo
\(P=\sum\dfrac{1}{x+y+1}\ge\dfrac{9}{2\left(x+y+z\right)+3}=\dfrac{9}{2.1+3}=\dfrac{9}{5}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Lời giải:
Do $x+y=1$ nên:
$P=\frac{x}{\sqrt{x+y-x}}+\frac{y}{\sqrt{x+y-y}}=\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}$
$=\frac{x^2}{x\sqrt{y}}+\frac{y^2}{y\sqrt{x}}$
$\geq \frac{(x+y)^2}{x\sqrt{y}+y\sqrt{x}}=\frac{1}{x\sqrt{y}+y\sqrt{x}}$ (áp dụng BĐT Cauchy-Schwarz)
Áp dụng BĐT Bunhiacopxky:
$(x\sqrt{y}+y\sqrt{x})^2\leq (x+y)(xy+xy)=2xy(x+y)\leq \frac{(x+y)^2}{2}(x+y)=\frac{1}{2}$
$\Rightarrow x\sqrt{y}+y\sqrt{x}\leq \frac{\sqrt{2}}{2}$
$\Rightarrow P\geq \frac{1}{x\sqrt{y}+y\sqrt{x}}\geq \frac{1}{\frac{\sqrt{2}}{2}}=\sqrt{2}$
Vậy $P_{\min}=\sqrt{2}$. Giá trị này đạt tại $x=y=\frac{1}{2}$.
\(2P=2x^2+8y^2+\dfrac{150}{x}+\dfrac{2}{y}\)
\(=\dfrac{7}{5}x^2+7y^2+\left(\dfrac{3}{5}x^2+\dfrac{75}{x}+\dfrac{75}{x}\right)+\left(y^2+\dfrac{1}{y}+\dfrac{1}{y}\right)\)
Ta có: \(\left(5+1\right)\left(x^2+5y^2\right)\ge5\left(x+y\right)^2\Rightarrow\dfrac{7\left(x^2+5y^2\right)}{5}\ge\dfrac{7\left(x+y\right)^2}{6}\ge42\)
\(\Rightarrow2P\ge42+3\sqrt[3]{\dfrac{3.75^2.x^2}{5x^2}}+3\sqrt[3]{\dfrac{y^2}{y^2}}=90\)
\(\Rightarrow P\ge45\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(5;1\right)\)
\(P=3x+\dfrac{12}{x}+y+\dfrac{16}{y}+2\left(x+y\right)\ge2\sqrt{3x.\dfrac{12}{x}}+2\sqrt{y.\dfrac{16}{y}}+2.6=32\)
\(\Rightarrow P_{min}=32\) khi \(\left\{{}\begin{matrix}3x=\dfrac{12}{x}\\y=\dfrac{16}{y}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)