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Q=\(\left(1+\dfrac{a}{x}\right)\left(1+\dfrac{a}{y}\right)\left(1+\dfrac{a}{z}\right)\)
\(Q=\left(\dfrac{x+a}{x}\right)\left(\dfrac{y+a}{y}\right)\left(\dfrac{z+a}{z}\right)\)\
=\(\left(\dfrac{2x+y+z}{x}\right)\left(\dfrac{2y+x+z}{y}\right)\left(\dfrac{2z+x+y}{z}\right)\)
=\(\dfrac{\left(2x+y+z\right)\left(2y+x+z\right)\left(2z+x+y\right)}{xyz}\)
ÁP dụng BĐT cô si
\(2x+y+z=x+x+y+z\ge4\sqrt[4]{x^2yz}\)
\(2y+x+z=y+y+x+z\ge4\sqrt[4]{y^2xy}\)
\(2z+y+x=z+z+x+y\ge4\sqrt[4]{z^2xy}\)
=> Q\(\ge\dfrac{64.\sqrt[4]{x^4y^4z^4}}{xyz}=64\)
=> MinQ=64 khi x=y=z=a/3
\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)
\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)
\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)
\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1^2}{\sqrt{a}}+\frac{2^2}{\sqrt{b}}+\frac{3^2}{\sqrt{c}}\ge\frac{\left(1+2+3\right)^2}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\frac{36}{6}=6\)
Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}\frac{1}{\sqrt{a}}=\frac{2}{\sqrt{b}}=\frac{3}{\sqrt{c}}\\\sqrt{a}+\sqrt{b}+\sqrt{c}=6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a}=1\\\sqrt{b}=2\\\sqrt{c}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=4\\c=9\end{matrix}\right.\)
từ giả thiết ,ta có:\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=4\)\(\Leftrightarrow a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1\)---> thay 1= vào ...
\(A=\dfrac{\left(a+b+c+a\right)\left(a+b+c+b\right)\left(a+b+c+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(A\ge\dfrac{2\sqrt{\left(a+b\right)\left(a+c\right)}.2\sqrt{\left(a+b\right)\left(b+c\right)}.2\sqrt{\left(a+c\right)\left(b+c\right)}}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=8\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Bài này hay:)
c = min {a,b,c}. Đặt
\(a-c=x;b-c=y\Rightarrow x,y\ge0\) và x + y = a + b - 2c \(=3-3c\le3\)
\(\Rightarrow a-b=x-y;c=\frac{3-x-y}{3}\)
\(a=x+c=x+\frac{3-x-y}{3}=\frac{2x-y+3}{3}\)
\(b=y+c=\frac{2y-x+3}{3}\)
Như vậy: \(K=\sqrt{4\left(2x-y+3\right)+y^2}+\sqrt{4\left(2y-x+3\right)+x^2}+\sqrt{4\left(3-x-y\right)+\left(x-y\right)^2}\)
\(=\sqrt{y^2-4y+8x+12}+\sqrt{x^2-4x+8y+12}+\sqrt{4\left(3-x-y\right)+\left(x-y\right)^2}\)
Giờ em đang bận, tối em làm tiếp!
\(12a+\left(b-c\right)^2=4a\left(a+b+c\right)+b^2-2bc+c^2\)
\(=4a^2+b^2+c^2+4ab+4ac+2bc-4bc\)
\(=\left(2a+b+c\right)^2-4bc\le\left(2a+b+c\right)^2\)
\(\Rightarrow\sqrt{12a+\left(b-c\right)^2}\le2a+b+c\)
Tương tự: \(\sqrt{12b+\left(a-c\right)^2}\le a+2b+c\); \(\sqrt{12c+\left(a-b\right)^2}\le a+b+2c\)
Cộng vế với vế:
\(K\le4\left(a+b+c\right)=12\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;3\right)\) và các hoán vị