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\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)
\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)
\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)
\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)
\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)
Dấu = xảy ra khi \(x=y=z=9\)
Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\)
CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\) ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)
\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
Mặt khác : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)
Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)
" = " \(\Leftrightarrow x=y=z=9\)
Ta có \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\left(x,y,z>0\right)\).
\(\Leftrightarrow\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\).
\(P=\frac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+y^2}\right)\)\(\left(x,y,z>0\right)\).
Ta có:
\(\sqrt{2y^2+2yz+2z^2}=\sqrt{\frac{5}{4}\left(y^2+2yz+z^2\right)+\frac{3}{4}\left(y^2-2yz+z^2\right)}\)
\(=\sqrt{\frac{5}{4}\left(y+z\right)^2+\frac{3}{4}\left(y-z\right)^2}\).
Ta có:
\(\frac{3}{4}\left(y-z\right)^2\ge0\forall y;z>0\).
\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2\ge\frac{5}{4}\left(y+z\right)^2\forall y;z>0\).
\(\Rightarrow\sqrt{\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y,z>0\).
\(\Leftrightarrow\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y;z>0\).
\(\Leftrightarrow x\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}x\left(y+z\right)\forall x;y;z>0\left(1\right)\).
Chứng minh tương tự, ta được:
\(y\sqrt{2x^2+xz+2z^2}\ge\frac{\sqrt{5}}{2}y\left(x+z\right)\forall x;y;z>0\left(2\right)\).
Chứng minh tương tự, ta được:
\(z\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}z\left(x+y\right)\forall x;y;z>0\left(3\right)\).
Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:
\(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+2y^2}\)\(\ge\)\(\frac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]=\sqrt{5}\left(xy+yz+zx\right)\).
\(\Leftrightarrow\frac{1}{xyz}\left(x\sqrt{2y^2+yz+z^2}+y\sqrt{2z^2+zx+2x^2}+z\sqrt{2x^2+xy+2y^2}\right)\)\(\ge\)\(\frac{\sqrt{5}\left(xy+yz+zx\right)}{xyz}=\sqrt{5}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\).
\(\Leftrightarrow P\ge\frac{\sqrt{5}}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\)
\(\left(4\right)\).
Vì \(x,y,z>0\)nên áp dụng bất đẳng thức Bu-nhi-a-cốp-xki, ta được:
\(\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\)\(\left(1.\frac{1}{\sqrt{x}}+1.\frac{1}{\sqrt{y}}+1.\frac{1}{\sqrt{z}}\right)^2\).
\(\Leftrightarrow\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2=1^2=1\)
(vì\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\)).
\(\Leftrightarrow\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\frac{\sqrt{5}}{3}\)\(\left(5\right)\).
Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:
\(P\ge\frac{\sqrt{5}}{3}\).
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}x=y=z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\end{cases}}\Leftrightarrow x=y=z=9\).
Vậy \(minP=\frac{\sqrt{5}}{3}\Leftrightarrow x=y=z=9\).
Ta có x2-xy+y2=\(\left(\dfrac{x+y}{2}\right)^2+3\left(\dfrac{x-y}{2}\right)^2\)\(\ge\)\(\left(\dfrac{x+y}{2}\right)^2\)
=>\(\dfrac{\sqrt{x^2-xy+y^2}}{x+y+2z}\ge\dfrac{x+y}{2\left(x+y+2z\right)}\)(1) . Tương tự ...
Đặt \(\left\{{}\begin{matrix}y+z=a\\x+z=b\\x+y=c\end{matrix}\right.\)(a,b,c>0). Khi đó ta có :
S=\(\dfrac{1}{2}\left(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\right)\ge\dfrac{3}{4}\) (Netbit)
Áp dụng BĐT Cô-si ta có:
\(2x^2+3xy+4y^2\ge3\sqrt[3]{2x^2\cdot3xy\cdot4y^2}=3\sqrt[3]{24x^3y^3}\Rightarrow\sqrt{2x^2+3xy+4y^2}\ge\sqrt{xy\cdot3\sqrt[3]{24}}\)
Tương tự: \(\sqrt{2y^2+3yz+4z^2}\ge\sqrt{yz\cdot3\sqrt[3]{24}}\); \(\sqrt{2z^2+3zx+4x^2}\ge\sqrt{zx\cdot3\sqrt[3]{24}}\)
Cộng theo vế 3 BĐT vừa tìm, ta được:
\(P\ge\sqrt{3\sqrt[3]{24}}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=\sqrt{3\sqrt[3]{24}}=\sqrt[6]{648}\)
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)
Ta có:
\(2x^2+xy+2y^2=x^2+y^2+\frac{3}{4}\left(x+y\right)^2+\frac{1}{4}\left(x-y\right)^2\)
\(\ge\frac{2\left(x+y\right)^2}{4}+\frac{3\left(x+y\right)^2}{4}=\frac{5\left(x+y\right)^2}{4}\)
\(\Rightarrow\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}\left(x+y\right)\). Tương tự ta có:
\(\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right);\sqrt{2z^2+xz+2x^2}\ge\frac{\sqrt{5}}{2}\left(x+z\right)\)
\(\Rightarrow M\ge\frac{\sqrt{5}}{2}\left(x+y\right)+\frac{\sqrt{5}}{2}\left(y+z\right)+\frac{\sqrt{5}}{2}\left(x+z\right)\)
\(=\sqrt{5}\left(x+y+z\right)=\sqrt{5}\)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)
Cho mình hối tại sao đẳng thức sảy ra x=y=z=1/3 vậy
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 \(\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}x^2+\frac{1}{2}xy+\frac{1}{4y^2}+\frac{3}{4}x^2-\frac{3}{2}xy+\frac{3}{4}y^2}\)
\(=\sqrt{\frac{1}{4}\left(x^2+2xy+y^2\right)+\frac{3}{4}\left(x^2-2xy+y^2\right)}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\)
Ta chứng minh điều sau: Nếu \(a,b>0\) thì \(2a^2+ab+2b^2\ge\frac{5\left(a+b\right)^2}{4}.\) Thực vậy bất đẳng thức cần chứng minh tương đương với
\(8a^2+4ab+8b^2\ge5\left(a^2+2ab+b^2\right)\Leftrightarrow3\left(a^2-2ab+b^2\right)\ge0\Leftrightarrow3\left(a-b\right)^2\ge0.\)
Quay lại bài toán, áp dụng nhận xét ta được
\(\sqrt{2x^2+xy+2y^2}\ge\frac{5\left(x+y\right)}{2},\sqrt{2y^2+yz+2z^2}\ge\frac{5\left(y+z\right)}{2},\sqrt{2z^2+zx+2x^2}\ge\frac{5\left(z+x\right)}{2}.\)
Cộng các bất đẳng thức lại ta sẽ được \(VT\ge\frac{5}{2}>\sqrt{5}.\)
hình như thiếu cái gì đó
Đề bài đủ rồi bạn nhé.