Áp dụng bất đẳng thức Cauchy-Swartz, ta có : \(P^2=\left(1.\sqrt{x-3}+1.\sqrt{y-4}\right)^2\le\left(1^2+1^2\right)\left(x-3+y-4\right)=2\left(x+y-7\right)\)

\(\Rightarrow P^2\le2\) (vì x+y=8)

\(\Rightarrow P\le\sqrt{2}\) . Dấu đẳng thức xảy ra <=> \(\begin{cases}x\ge3;y\ge4\\x+y=8\\\sqrt{x-3}=\sqrt{y-4}\end{cases}\Leftrightarrow\begin{cases}x=\frac{7}{2}\\y=\frac{9}{2}\end{cases}\)

Vậy Max P = \(\sqrt{2}\Leftrightarrow\begin{cases}x=\frac{7}{2}\\y=\frac{9}{2}\end{cases}\)

Đặt \(\left\{{}\begin{matrix}x=sina\\y=sinb\end{matrix}\right.\) với \(a;b\in\left(0;\dfrac{\pi}{2}\right)\)

\(P=\sqrt{sina}+\sqrt{sinb}+\sqrt[4]{12}.\sqrt{sina.cosb+cosa.sinb}\)

\(P\le\sqrt{2\left(sina+sinb\right)}+\sqrt[4]{12}.\sqrt{sin\left(a+b\right)}\)

Do \(sina+sinb=2sin\dfrac{a+b}{2}cos\dfrac{a-b}{2}\le2sin\dfrac{a+b}{2}\)

\(\Rightarrow P\le2\sqrt{sin\dfrac{a+b}{2}}+\sqrt[4]{12}.\sqrt{sin\left(a+b\right)}=2\sqrt{sint}+\sqrt[4]{12}.\sqrt{sin2t}\)

\(\Rightarrow\dfrac{P}{\sqrt{2}}\le\sqrt{2sint}+\sqrt{\sqrt{3}.sin2t}\Rightarrow\dfrac{P^2}{4}\le2sint+\sqrt{3}sin2t\)

\(\Rightarrow\dfrac{P^2}{8}\le sint\left(1+\sqrt{3}cost\right)\Rightarrow\dfrac{P^4}{64}\le sin^2t\left(1+\sqrt{3}cost\right)^2\le2sin^2t\left(1+3cos^2t\right)\)

\(\Leftrightarrow\dfrac{P^4}{128}\le sin^2t\left(4-3sin^2t\right)=-3sin^4t+4sin^2t\)

\(\Leftrightarrow\dfrac{P^4}{128}\le-3\left(sin^2t-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\le\dfrac{4}{3}\)

\(\Rightarrow P\le4.\sqrt[4]{\dfrac{2}{3}}\)

Dấu "=" xảy ra khi và chỉ khi \(sint=\sqrt{\dfrac{2}{3}}\)

Cách khác:

Đk: \(x\ge2,y\ge3\)

Với a,b\(\ge\) 0có:

\(a+b\le\sqrt{2\left(a^2+b^2\right)}\) <=> \(a^2+2ab+b^2\le2a^2+2b^2\) <=> \(0\le a^2-2ab+b^2\)

<=>\(0\le\left(a-b\right)^2\)

Dấu "=" xảy ra <=>a=b>0

Áp dụng bđt trên có:

\(S=\sqrt{x-2}+\sqrt{y-3}\le\sqrt{2\left(x-2+y-3\right)}=\sqrt{2\left(6-2-3\right)}\)(do x+y=6)

=> \(S\le\sqrt{2}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{y-3}\\x+y=6\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x-2=y-3\\x+y=6\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y-x=1\\x+y=6\end{matrix}\right.\) <=> x=2,5 và y=3,5(t/m)

\(S^2\le\left(1+1\right)\left(x-2+y-3\right)=2\left(x+y-5\right)=2\)

\(\Rightarrow S\le\sqrt{2}\)

\(\Rightarrow S_{max}=\sqrt{2}\) khi \(\left\{{}\begin{matrix}x-2=y-3\\x+y=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=\frac{7}{2}\end{matrix}\right.\)

ĐKXĐ :\(x\ge2;y\ge3\)

\(S^2=x-2+y-3+2\sqrt{\left(x-2\right)\left(y-3\right)}\)

\(\ge6-2-3+0=1\)

Vì \(S>0\Rightarrow S\ge1\)

Vậy \(Max_S=1\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=3\end{matrix}\right.\end{matrix}\right.\)

\(3=x+y+xy\le\sqrt{2\left(x^2+y^2\right)}+\dfrac{x^2+y^2}{2}\)

\(\Rightarrow\left(\sqrt{x^2+y^2}-\sqrt{2}\right)\left(\sqrt{x^2+y^2}+3\sqrt{2}\right)\ge0\)

\(\Rightarrow x^2+y^2\ge2\)

\(\Rightarrow-\left(x^2+y^2\right)\le-2\)

\(P=\sqrt{9-x^2}+\sqrt{9-y^2}+\dfrac{x+y}{4}\le\sqrt{2\left(9-x^2+9-y^2\right)}+\dfrac{\sqrt{2\left(x^2+y^2\right)}}{4}\)

\(P\le\sqrt{2\left(18-x^2-y^2\right)}+\dfrac{1}{4}.\sqrt{2\left(x^2+y^2\right)}\)

\(P\le\left(\sqrt{2}-1\right)\sqrt{18-x^2-y^2}+\sqrt[]{2}\sqrt{\dfrac{\left(18-x^2-y^2\right)}{2}}+\dfrac{1}{2}\sqrt{\dfrac{x^2+y^2}{2}}\)

\(P\le\left(\sqrt{2}-1\right).\sqrt{18-2}+\sqrt{\left(2+\dfrac{1}{4}\right)\left(\dfrac{18-x^2-y^2+x^2+y^2}{2}\right)}=\dfrac{1+8\sqrt{2}}{2}\)

Dấu "=" xảy ra khi \(x=y=1\)

Tham khảo:

Cho 3 số thức x,y,z thỏa mãn \(x\ge1;y\ge4;z\ge9\) tìm giá trị lớn nhất của biết thức Q=\(\dfrac{yz\sqrt{x-1}+zx\sqrt... - Hoc24

a) ĐKXĐ: \(x,y\ge0\)

\(M=\dfrac{x\sqrt{y}-\sqrt{y}-y\sqrt{x}+\sqrt{x}}{1+\sqrt{xy}}=\dfrac{x\sqrt{y}-y\sqrt{x}+\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\)

\(=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\left(\sqrt{x}-\sqrt{y}\right)}{1+\sqrt{xy}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{1+\sqrt{xy}}=\sqrt{x}-\sqrt{y}\)

b) \(x=\left(1-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)

\(y=3-\sqrt{8}\Rightarrow\sqrt{y}=\sqrt{3-\sqrt{8}}=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)

\(\Rightarrow M=\left(\sqrt{3}-1\right)-\left(\sqrt{2}-1\right)=\sqrt{3}-\sqrt{2}\)

2. \(P=x^2-x\sqrt{3}+1=\left(x^2-x\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=\left(x-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)

Dấu '=' xảy ra khi \(x=\frac{\sqrt{3}}{2}\)

Vây \(P_{min}=\frac{1}{4}\)khi \(x=\frac{\sqrt{3}}{2}\)

3. \(Y=\frac{x}{\left(x+2011\right)^2}\le\frac{x}{4x.2011}=\frac{1}{8044}\)

Dấu '=' xảy ra khi \(x=2011\)

Vây \(Y_{max}=\frac{1}{8044}\)khi \(x=2011\)

4. \(Q=\frac{1}{x-\sqrt{x}+2}=\frac{1}{\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{7}{4}}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}\le\frac{4}{7}\)

Dấu '=' xảy ra khi \(x=\frac{1}{4}\) 

Vậy \(Q_{max}=\frac{4}{7}\)khi \(x=\frac{1}{4}\)

Làm như thế nào ra \(\frac{x}{4x.2011}\)vậy bạn?

BĐT \(\left(x+y\right)^2\ge4xy\)nhe bạn