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1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:
\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).
Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).
2.
\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)
Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)
\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )
\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)
\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)
Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)
3. Chia 2 vế giả thiết cho \(x^2y^2\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)
\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
Bài 1 : x = 0 ; y = 2
Bài 2 Max A = 1 <=> x = 0 , y = 1 hoặc x = 1 , y = 0
Min A = 0,5 <=> x = y = 0,5
Bài 1:
ĐK: \(x,y\ge-2\)
Ta có: \(\sqrt{x+2}-y^3=\sqrt{y+2}-x^3\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+\frac{x-y}{\sqrt{x+2}+\sqrt{y+2}}=0\)
=> x-y=0=>x=y
Thay y=x vào B ta được: B=x2+2x+10\(=\left(x+1\right)^2+9\ge9\forall x\ge-2\)
Dấu '=' xảy ra <=> x+1=0=>x=-1 (tmđk)
Vậy Min B =9 khi x=y=-1
Ta có:
\(\left(x-y\right)^2+\left(x-z\right)^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(x+y+z\right)^2\ge\left(x+y+z\right)^2\)
\(\Leftrightarrow x^2-2xy+y^2+x^2-2xz+z^2+x^2+y^2+z^2+2\left(xy+yz+xz\right)\ge A^2\)
\(\Leftrightarrow A^2\le2\left(y^2+yz+z^2\right)+3x^2=36\)
\(\Leftrightarrow-6\le A\le6\)
\(2x^2+7x+7y+2xy+y^2+12=0\)
\(\Leftrightarrow\left(x^2+y^2+4+2\left(xy+2x+2y\right)\right)+3\left(x+y+2\right)+2=-x^2\)
\(\Leftrightarrow\left(x+y+2\right)^2+3\left(x+y+2\right)+2=-x^2\)
\(\Leftrightarrow P^2+3P+2=-x^2\le0\)
\(\Leftrightarrow-2\le P\le-1\)
+) \(P=\sqrt{7x+9}+\sqrt{7y+9}+\sqrt{7z+9}\)
\(P^2\le3\left(7x+7y+7z+27\right)=102\)
\(P\le\sqrt{102}\)
\(MaxP=102\Leftrightarrow x=y=z=\dfrac{1}{3}\)
+) \(x,y,z\in[0;1]\)\(\Rightarrow\left\{{}\begin{matrix}x\ge x^2\\y\ge y^2\\z\ge z^2\end{matrix}\right.\)
\(P\ge\sqrt{x^2+6x+9}+\sqrt{y^2+6y+9}+\sqrt{z^2+6z+9}\)
\(=x+y+z+9=10\)
\(MinP=10\Leftrightarrow\left(x;y;z\right)=\left(0;0;1\right)\text{và các hoán vị}\)
