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Ta co : \(\dfrac{2\sqrt{x.\left(x-z\right)}}{2}\le\dfrac{x+x-z}{2}\)
\(\dfrac{2\sqrt{z\left(y-x\right)}}{2}\le\dfrac{z+y-x}{2}\)
VT≤\(\dfrac{2x-z}{2}+\dfrac{z+y-x}{2}=\dfrac{2x-z+z+y-x}{2}\)
=\(\dfrac{x+y}{2}\le\sqrt{xy}\)
=> DPCM
Toan bo dung bdt Co Si nha
(x+y+z)^2/3>=(x^2+y^2+z^2+2xy+2yz+2zx)/3>=3(xy+yz+zx)/3=xy+yz+zx(do x^2+y^2+z^2>=xy+yz+zx)(1)
(xy+yz)/2>=y√xz;(yz+zx)/2>=z√xy;(zx+xy)/2>=x√yz(BĐT Cô-si)
Cộng theo vế >>>xy+yz+zx>=y√xz+z√xy+x√yz(2)
Từ(1),(2) >>>đpcm
3, \(P=a+b+\frac{1}{2a}+\frac{2}{b}\)
=\(\left(\frac{1}{2a}+\frac{a}{2}\right)+\left(\frac{b}{2}+\frac{2}{b}\right)+\frac{a+b}{2}\)
AD bđt cosi vs hai số dương có:
\(\frac{1}{2a}+\frac{a}{2}\ge2\sqrt{\frac{1}{2a}.\frac{a}{2}}=2\sqrt{\frac{1}{4}}=1\)
\(\frac{b}{2}+\frac{2}{b}\ge2\sqrt{\frac{b}{2}.\frac{2}{b}}=2\)
Có \(\frac{a+b}{2}\ge\frac{3}{2}\) (vì a+b \(\ge3\))
=> \(P=\left(\frac{1}{2a}+\frac{a}{2}\right)+\left(\frac{b}{2}+\frac{2}{b}\right)+\frac{a+b}{2}\ge1+2+\frac{3}{2}\)
<=> P \(\ge4.5\)
Dấu "=" xảy ra <=>\(\left\{{}\begin{matrix}\frac{1}{2a}=\frac{a}{2}\\\frac{b}{2}=\frac{2}{b}\\a+b=3\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a^2=1\\b^2=4\\a+b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=1\\b=2\\a+b=3\end{matrix}\right.\)
=> a=2,b=3
Vậy minP=4.5 <=>a=1,b=2
Ta có
\(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left[4\left(4-y-z\right)+yz\right]}\)
\(=\sqrt{x\left(4\left(x+\sqrt{xyz}\right)+yz\right)}\)
\(=\sqrt{4x^2+4x\sqrt{xyz}+xyz}\)
\(=2x+\sqrt{xyz}\)
Khi đó \(T=2\left(x+y+z\right)+3\sqrt{xyz}-\sqrt{xyz}=2.4=8\)
A
Áp dụng BĐT cosi ta có
\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)
\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)
Khi đó
\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)
MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)
B
Áp dụng BĐT cosi ta có :
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)
Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)
=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)
\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z