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\(A=x-2y+3z\left(x,y,z>0\right)\)
\(\left\{{}\begin{matrix}2x+4x+3z=8\left(1\right)\\3x+y-3z=2\left(2\right)\end{matrix}\right.\)
(1) <=> \(5x+5y=10\) <=> x+ y = 2
=> y = 2-x
Từ (1) => \(2x+4\left(2-x\right)+3z=8\)
=> -2x +3z =0
=> \(x=\dfrac{3}{2}z\) => \(z=\dfrac{2}{3}x\) thay vào A
=> \(A=x-2\left(2-x\right)+3.\dfrac{2}{3}x=5x-4\ge-4\)
Vậy Amin = -4.
Cho x,y,z >0 thỏa mãn x+y+z = 2. Tìm GTLN của biểu thức
\(P=\sqrt{2x+yz}+\sqrt{2y+xz}+\sqrt{2z+xy}\)
\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{1}{2}\left(x+y+x+z\right)=\dfrac{1}{2}\left(2x+y+z\right)\)
Tương tự: \(\sqrt{2y+xz}\le\dfrac{1}{2}\left(x+2y+z\right)\) ; \(\sqrt{2z+xy}\le\dfrac{1}{2}\left(x+y+2z\right)\)
Cộng vế:
\(P\le\dfrac{1}{2}\left(4x+4y+4z\right)=4\)
\(P_{max}=4\) khi \(x=y=z=\dfrac{2}{3}\)
P = \(1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)
\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)
\(=\sqrt{3.\left(4+xy+yz+zx\right)}\)
Đã biết x2 + y2 + z2 \(\ge\)xy + yz + zx
=> xy + yz + zx \(\le\dfrac{\left(x+y+z\right)^2}{3}\)
Khi đó \(P\le\sqrt{3\left(4+xy+yz+zx\right)}\le\sqrt{3\left[4+\dfrac{\left(x+y+z\right)^2}{3}\right]}\)
= 4
Dấu "=" xảy ra <=> x = 2/3
\(\sqrt{2x+yz}=\sqrt{\left(x+y+z\right)x+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{x+2y+z}{2}\\ \Leftrightarrow P=\sum\sqrt{2x+yz}\le\dfrac{x+2y+z+2x+y+z+x+y+2z}{2}=\dfrac{4\left(x+y+z\right)}{2}=2\cdot2=4\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{2}{3}\)
Cho ba số thực dương x;y;z thoả mãn \(5\left(x+y+z\right)^2\ge14\left(x^2+y^2+z^2\right)\) Tìm giá trị lớn nhất nhỏ nh... - Hoc24
\(T=\dfrac{\left(xy\right)^2}{zx+zy}+\dfrac{\left(yz\right)^2}{xy+xz}+\dfrac{\left(zx\right)^2}{yx+yz}\ge\dfrac{xy+yz+zx}{2}\ge\dfrac{3}{2}\sqrt[3]{\left(xyz\right)^2}=\dfrac{3}{2}\)
Chắc đề là \(x+y+z=3\)
Ta có:
\(\left(2x+y+z\right)^2=\left(x+y+x+z\right)^2\ge4\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow P\le\dfrac{x}{4\left(x+y\right)\left(x+z\right)}+\dfrac{y}{4\left(x+y\right)\left(y+z\right)}+\dfrac{z}{4\left(x+z\right)\left(y+z\right)}\)
\(\Rightarrow P\le\dfrac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{4\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\dfrac{xy+yz+zx}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Mặt khác:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(zy+yz+zx\right)=\dfrac{8}{3}\left(xy+yz+zx\right)\)
\(\Rightarrow P\le\dfrac{xy+yz+zx}{2.\dfrac{8}{3}\left(xy+yz+zx\right)}=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
\(x^2y^2+1\ge2xy;y^2z^2+1\ge2yz;z^2x^2+1\ge2zx\)
\(\Rightarrow x^2y^2+y^2z^2+z^2x^2\ge2\left(xy+yz+zx\right)-3\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)-3\le6\)
\(\Rightarrow\left(x+y+z\right)^2-3\le6\)
\(\Rightarrow x+y+z\le3\)
Dấu " = " xảy ra khi \(x=y=z=1\)
