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Ta có : 2P = \(\frac{\sqrt{4x^2-4xy+4y^2}}{x+y+2z}+\frac{\sqrt{4y^2-4yz+4z^2}}{y+z+2x}+\frac{\sqrt{4z^2-4zx+4x^2}}{z+x+2y}\)
\(=\frac{\sqrt{\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2}}{x+y+2z}+\frac{\sqrt{\left(2y-z\right)^2+\left(\sqrt{3}z\right)^2}}{y+z+2x}+\frac{\sqrt{\left(2z-x\right)^2+\left(\sqrt{3}x\right)^2}}{z+x+2y}\)
Lại có \(\frac{\sqrt{\left[\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2\right]\left[\left(1^2+\left(\sqrt{3}\right)^2\right)\right]}}{x+y+2z}\ge\frac{\left[\left(2x-y\right).1+3y\right]}{x+y+2z}=\frac{2\left(x+y\right)}{x+y+2z}\)
=> \(\sqrt{\frac{\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2}{x+y+2z}}\ge\frac{x+y}{x+y+2z}\)(BĐT Bunyakovsky)
Tương tự ta đươc \(2P\ge\frac{x+y}{x+y+2z}+\frac{y+z}{2x+y+z}+\frac{z+x}{2y+z+x}\)
Đặt x + y = a ; y + z = b ; x + z = c
Khi đó \(2P\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)
\(\ge\left(a+b+c\right).\frac{9}{2\left(a+b+c\right)}-3\ge\frac{9}{2}-3=\frac{3}{2}\)
=> \(P\ge\frac{3}{4}\)
Dấu "=" xảy ra <=> x = y = z
bài 8 : bỏ dấu hoặc rồi tính
a;( 17 - 299) + ( 17 - 25 + 299)
Đây là câu bđt của chuyên Quảng Nam vừa thi mà:vvv
Ta có: \(xy+yz+zx=xyz\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\left(a,b,c>0\right)\)
Khi đó: \(H=\frac{a}{9b^2+1}+\frac{b}{9c^2+1}+\frac{c}{9a^2+1}\)
\(=\left(a+b+c\right)-\left(\frac{9ab^2}{9b^2+1}+\frac{9bc^2}{9c^2+1}+\frac{9ca^2}{9a^2+1}\right)\)
\(\ge1-\left(\frac{9ab^2}{6b}+\frac{9bc^2}{6c}+\frac{9ca^2}{6a}\right)\)
\(=1-\frac{3}{2}\left(ab+bc+ca\right)\ge1-\frac{3}{2}\cdot\frac{\left(a+b+c\right)^2}{3}=1-\frac{3}{2}\cdot\frac{1}{3}=\frac{1}{2}\)
Dấu "=" xảy ra khi: \(x=y=z=3\)
Vậy Min(H) = 1/2 khi x = y = z = 3
\(P=\frac{y^3}{x^2+xy+y^2}+\frac{z^3}{y^2+zx+z^2}+\frac{x^3}{z^2+zx+x^2}\)
\(\Leftrightarrow P=\frac{y^4}{x^2y+xy^2+y^3}+\frac{z^4}{y^2z+z^2x+z^3}+\frac{x^4}{z^2x+zx^2+x^3}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^3+y^3+z^3+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}\)
\(\Leftrightarrow P\ge\frac{\left(x^2+y^2+z^2\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2\right)}=\frac{x^2+y^2+z^2}{x+y+z}\ge3\)
Dấu "=" khi x=y=z=3
ta có \(\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\ge\sqrt{\frac{1}{4}\left(x+y\right)^2}=\frac{1}{2}\left(x+y\right)\)
tương tự ta có các trường hợp còn lại và ta có
\(S\ge\frac{1}{2}\left(\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{z+x}{z+x+2y}\right)\)
đặt \(x+y=a;y+z=b;z+x=c\)
=> \(S\ge\frac{1}{2}\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
đặt \(A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ca+ca}\)
Áp dụng bđt svác sơ ta có
\(A\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
mạt khác Áp dụng bđt cô si ta có
\(\hept{\begin{cases}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{cases}}\)
=> \(a^2+b^2+c^2\ge2\left(ab+bc+ca\right)\)
=> \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
=> \(A\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
=> \(S\ge\frac{3}{4}\)
dấu = xảy ra <=> x=y=z>o
ta có \(\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}x^2+\frac{1}{2}xy+\frac{1}{4y^2}+\frac{3}{4}x^2-\frac{3}{2}xy+\frac{3}{4}y^2}\)
\(=\sqrt{\frac{1}{4}\left(x^2+2xy+y^2\right)+\frac{3}{4}\left(x^2-2xy+y^2\right)}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\)
a/
Với mọi số thực x;y;z ta luôn có:
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2yz\ge3xy+3yz+3zx\)
\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\) (đpcm)
b/
\(M=2\left(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}\right)+\frac{1}{xy+yz+zx}\)
\(M\ge2.\frac{9}{x^2+y^2+z^2+xy+yz+zx+xy+yz+zx}+\frac{1}{\frac{\left(x+y+z\right)^2}{3}}\)
\(M\ge\frac{18}{\left(x+y+z\right)^2}+\frac{3}{\left(x+y+z\right)^2}=\frac{21}{\left(x+y+z\right)^2}=21\)
\(M_{min}=21\) khi \(x=y=z=\frac{1}{3}\)
\(T\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+x+y+z}=\dfrac{x+y+z}{2}\ge\dfrac{2019}{2}\)
áp dụng BĐT:\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\) với a,b,c,x,y,z là số dương
ta có BĐT Bunhiacopxki cho 3 bộ số:\(\left(\dfrac{a}{\sqrt{x}};\sqrt{x}\right);\left(\dfrac{b}{\sqrt{y}};\sqrt{y}\right);\left(\dfrac{c}{\sqrt{z}};\sqrt{z}\right)\)
ta có :
\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\left(x+y+z\right)\)\(=\left[\left(\dfrac{a}{\sqrt{x}}\right)^2+\left(\dfrac{b}{\sqrt{y}}\right)^2+\left(\dfrac{c}{\sqrt{z}}\right)^2\right]\).\(\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]\)\(\ge\left(\dfrac{a}{\sqrt{x}}.\sqrt{x}+\dfrac{b}{\sqrt{y}}.\sqrt{y}+\dfrac{c}{\sqrt{z}}.\sqrt{z}\right)^2=\left(a+b+c\right)^2\)
lúc đó ta có :\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\)
ta có \(T=\dfrac{x^2}{x+\sqrt{yz}}+\dfrac{y^2}{y+\sqrt{zx}}+\dfrac{z^2}{z+\sqrt{xy}}\)\(\ge\dfrac{\left(x+y+z\right)^2}{x+\sqrt{yz}+y+\sqrt{zx}+z+\sqrt{xy}}\) mà ta có :
\(\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\)\(\le\dfrac{x+y}{2}+\dfrac{x+z}{2}+\dfrac{z+y}{2}\)\(\Rightarrow\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\le x+y+z\)
\(\Rightarrow T=\dfrac{2019}{2}\Leftrightarrow x=y=z=673\)
vậy \(\text{MinT}=\dfrac{2019}{2}\) khi và chỉ khi x=y=z=673
\(\Rightarrow\left(x+y+z\right)^2=1\Leftrightarrow x^2+y^2+z^2+2xy+2xz+2yz=1\)
\(P=\frac{3}{xy+xz+yz}+\frac{2}{x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{xy+xz+yz}+\frac{2\left(x^2+y^2+z^2\right)}{x^2+y^2+z^2}\)
\(P=\frac{3\left(x^2+y^2+z^2\right)+6\left(xy+xz+yz\right)}{xy+xz+yz}+\frac{2\left(x^2+y^2+z^2\right)+4\left(xy+xz+yz\right)}{x^2+y^2+z^2}\)
\(P=\frac{3\left(x^2+y^2+z^2\right)}{xy+xz+yz}+6+2+\frac{4\left(xy+xz+yz\right)}{x^2+y^2+z^2}\)
Áp dụng bất đẳng thức Cauchy ta có
\(\frac{3\left(x^2+y^2+z^2\right)}{xy+xz+yz}+\frac{4\left(xy+xz+yz\right)}{x^2+y^2+z^2}\ge2\sqrt{12}=4\sqrt{3}\)
\(\Rightarrow P\ge4\sqrt{3}+6+2=8+4\sqrt{3}\)
Dấu bằng thì bạn tự xét nhé
$\Rightarrow \left(x+y+z\right)^2=1\Leftrightarrow x^2+y^2+z^2+2xy+2xz+2yz=1$
$P=\frac{3}{xy+xz+yz}+\frac{2}{x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{xy+xz+yz}+\frac{2\left(x^2+y^2+z^2\right)}{x^2+y^2+z^2}$
$P=\frac{3\left(x^2+y^2+z^2\right)+6\left(xy+xz+yz\right)}{xy+xz+yz}+\frac{2\left(x^2+y^2+z^2\right)+4\left(xy+xz+yz\right)}{x^2+y^2+z^2}$
$P=\frac{3\left(x^2+y^2+z^2\right)}{xy+xz+yz}+6+2+\frac{4\left(xy+xz+yz\right)}{x^2+y^2+z^2}$
$\frac{3\left(x^2+y^2+z^2\right)}{xy+xz+yz}+\frac{4\left(xy+xz+yz\right)}{x^2+y^2+z^2}\ge2\sqrt{12}=4\sqrt{3}$