Cho x,y,z la cac so duong va x+y+z=1. Tim GTNN cua M=xy+yz+zx
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Cho x,y,z la cac so duong va x+y+z =1 .Tim GTLN cua M =xy+yz+zx
Cho x,y,z la cac so khong am va x+y+z=1.Tim GTLN cua M=xy+yz+zx
\(x+z+y=1\Leftrightarrow\left(x+y+z\right)^2=1\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2zx\ge3\left(xy+yz+zx\right)=1\Rightarrow M_{max}=\frac{1}{3}.\text{Dâu "=" xay ra }\Leftrightarrow x=y=z=\frac{1}{3}\)
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Đơn giản hơn:
Áp dụng bđt quen thuộc \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)
Ta có: \(M\le\frac{\left(x+y+z\right)^2}{3}=\frac{1}{3}\)
Đẳng thức xảy ra khi x = y = z =1/3
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cho x,y,z la cac so thuc duong thoa man x+y+z=1 tim min A=x^3/(x^2+xy+y^2)+y^3/(y^2+yz+z^2)+z^3/(z^2+zx+x^2)
cho x,y,z la cac so huu ti duong thoa man x+1/yz y +1/xz z+1/xy la cac so nguyen tim gia tri lon nhat cua bieu thuc A=x+y^2+z^3
cho x,y,z la cac so nguyen duong thoa man \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2015\)
tinh gia tri lon nhat cua bieu thuc P=\(\dfrac{xy}{x^3+y^3}+\dfrac{yz}{y^3+z^3}+\dfrac{zx}{z^{3+x^3}}\)
cho x,y,z>0 va thoa man x+y+z=1. Tim GTNN cua F= 14(x2 +y2 +z2 ) +\(\frac{xy+yz+zx}{x^2y+y^2z+z^2x}\)
Cho x,y,z la cac so thuc duong thoa man x + y + z = 6
Tim GTNN cua bieu thuc P = ( x + y )/(xyz)
\(P=\frac{x+y}{xyz}=\frac{x}{xyz}+\frac{y}{xyz}=\frac{1}{yz}+\frac{1}{xz}\)
Áp dụng Bunyakovsky dạng phân thức : \(\frac{1}{yz}+\frac{1}{xz}\ge\frac{4}{z\left(x+y\right)}\)(1)
Ta có : \(\sqrt{z\left(x+y\right)}\le\frac{x+y+z}{2}\)( theo AM-GM )
=> \(z\left(x+y\right)\le\left(\frac{x+y+z}{2}\right)^2=\left(\frac{6}{2}\right)^2=9\)
=> \(\frac{1}{z\left(x+y\right)}\ge\frac{1}{9}\)=> \(\frac{4}{z\left(x+y\right)}\ge\frac{4}{9}\)(2)
Từ (1) và (2) => \(P=\frac{x+y}{xyz}=\frac{1}{yz}+\frac{1}{xz}\ge\frac{4}{z\left(x+y\right)}\ge\frac{4}{9}\)
=> P ≥ 4/9
Vậy MinP = 4/9, đạt được khi x = y = 3/2 ; z = 3
cho A= x^2+y^2+z^2+xy+yz+zx .Tim GTNN cua A ,biet x+y+z=3
cho x,y,z la cac so thuc thoa x+y+z=0, x+1>0, y+1>0, z+1>0. tim GTLN cua P=\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+4}\)
cho x,y,z,t la cac so duong. tim GTNN cua A=\(\frac{x-t}{t+y}+\frac{t-y}{y+z}+\frac{y-z}{z+x}+\frac{z-x}{x+t}\)