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\(\sqrt{x\left(y+z\right)}\le\frac{x+y+z}{2}\)( Cauchy)
\(\Rightarrow\sqrt{\frac{x}{y+z}}=\frac{x}{\sqrt{x\left(y+z\right)}}\le\frac{x}{\frac{x+y+z}{2}}=\frac{2x}{x+y+z}\)
Chứng minh tương tự:
\(\sqrt{\frac{y}{x+z}}\le\frac{2y}{x+y+z};\sqrt{\frac{z}{x+y}}\le\frac{2z}{x+y+z}\)
Cộng theo vế suy ra đocn. Dấu "=" ko xảy ra
đặt A=x/x+y+z +y/y+z+t +z/z+t+x +t/t+x+y
ta có x/x+y+z>x/x+y+z+t
y/y+z+t>y/x+y+z+t
z/z+t+x>z/z+t+x+y
t/t+x+y>t/x+t+y+z
=>A>x/x+y+t+z +t/x+y+t+z +z/x+y+t+z +y/x+t+y+z=x+y+z+t/x+y+z+t=1>3/4 (1)
*)y/y+z+t<y+x/y+z+t+x
x/x+y+z<x+t/x+y+z+t
z/z+t+x<z+y/x+y+z+t
t/t+x+y<t+z/t+x+y+z
=>A<y+x/x+y+z+t +x+t/x+y+z+t +z+y/x+y+z+t +t+z/x+y+z+t
=y+x+x+t+z+y+t+z/x+y+z+t=2(x+y+z+t)/x+y+z+t=2<5/2 (2)
từ (1) và (2) =>3/4<A<5/2
=>
Ta có:
\(\frac{x}{x+y+z+t}+\frac{y}{x+y+z+t}+\frac{z}{x+y+z+t}+\frac{t}{x+y+z+t}<\frac{x}{x+y+z}+\frac{y}{y+z+t}+\frac{z}{z+t+x}+\frac{t}{t+x+y}<\frac{x+t}{x+y+z+t}+\frac{x+y}{x+y+z+t}+\frac{y+z}{x+y+z+t}+\frac{z+t}{x+y+z+t}\)
\(\Rightarrow1<\frac{x}{x+y+z}+\frac{y}{y+z+t}+\frac{z}{z+t+x}+\frac{t}{t+x+y}<2\)
\(\Rightarrow\frac{3}{4}<\frac{x}{x+y+z}+\frac{y}{y+z+t}+\frac{z}{z+t+x}+\frac{t}{t+x+y}<\frac{5}{2}\)
Áp dụng bđt cauchy schwarz dạng engel , ta có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=\frac{1^2}{x}+\frac{1^2}{y}+\frac{1^2}{z}+\frac{1^2}{t}\ge\frac{16}{x+y+z+t}\)
\(< =>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+1\ge\frac{16}{x+y+z+t}+1\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z=t\)
Vậy ta có điều phải chứng minh
cách khác :3
Áp dụng bđt phụ : \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(< =>\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(< =>\frac{a+b}{ab}.\left(a+b\right).ab\ge\frac{4}{a+b}.\left(a+b\right).ab\)
\(< =>\left(a+b\right)^2\ge4ab\)
\(< =>a^2+2ab+b^2\ge4ab\)
\(< =>\left(a-b\right)^2\ge\)(luôn đúng)
Nên ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+1\ge\frac{4}{x+y}+\frac{4}{z+t}+1\ge\frac{16}{x+y+z+t}+1\)
từ đó =>3x=y+z+t
=>4x=x+y+z+t
tương tự=>4y=x+y+z+t
4z=x+y+z+t
4t=x+y+z+t
=>x=y=z=t=>F=4
mà bài này lớp 7 chứ,có phải lớp 9 đâu
\(\frac{x^4+y^4+z^4+t^4}{x^3+y^3+z^3+t^3}=\frac{\left(x^4+y^4+z^4+t^4\right)\left(x^2+y^2+z^2+t^2\right)}{\left(x^3+y^3+z^3+t^3\right)\left(x^2+y^2+z^2+t^2\right)}\)
\(\ge\frac{x^3+y^3+z^3+t^3}{x^2+y^2+z^2+t^2}=\frac{\left(x^3+y^3+z^3+t^3\right)\left(x+y+z+t\right)}{\left(x^2+y^2+z^2+t^2\right)\left(x+y+z+t\right)}\)
\(\ge\frac{x^2+y^2+z^2+t^2}{x+y+z+t}\ge\frac{\left(x+y+z+t\right)^2}{4\left(x+y+z+t\right)}=\frac{1}{4}\)
Dấu "=" xảy ra tại x=y=z=t=1/4
Bài làm có tham khảo của GOD Đạt Hồ
Ta có :
\(A=\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}\)
\(\Rightarrow A=\frac{1+z+x^2}{1+y+z^2}+\frac{1+x+y^2}{1+z+x^2}+\frac{1+y+z^2}{1+x+y^2}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3\sqrt[3]{\frac{1+z+x^2}{1+y+z^2}.\frac{1+x+y^2}{1+z+x^2}.\frac{1+y+z^2}{1+x+y^2}}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{y+2z}+\frac{x}{z+2x}+\frac{y}{x+2y}\right)\)
\(\Rightarrow A\ge3-\left(\frac{1}{2}-\frac{y}{2\left(y+2z\right)}+\frac{1}{2}-\frac{z}{2\left(z+2x\right)}+\frac{1}{2}-\frac{x}{2\left(x+2y\right)}\right)\)
\(\Rightarrow A\ge3-\frac{3}{2}+\frac{1}{2}\left(\frac{y}{y+2z}+\frac{z}{z+2x}+\frac{x}{x+2y}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}+\frac{x^2}{x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{y^2+2yz+z^2+2xz+x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}\right)\)
\(\Rightarrow A\ge2\)
Dấu " = " xảy ra khi \(x=y=z=1\)
Ta có :
\(\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}\)
\(\Rightarrow A=\frac{1+z+x^2}{1+y+z^2}+\frac{1+x+y^2}{1+z+x^2}+\frac{1+y+z^2}{1+x+y^2}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3\sqrt[3]{\frac{1+z+x^2}{1+y+z^2}.\frac{1+x+y^2}{1+z+x^2}.\frac{1+y+z^2}{1+x+y^2}}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{y+2z}+\frac{x}{z+2x}+\frac{y}{x+2y}\right)\)
\(\Rightarrow A\ge3-\left(\frac{1}{2}-\frac{y}{2\left(y+2z\right)}+\frac{1}{2}-\frac{z}{2\left(z+2x\right)}+\frac{1}{2}-\frac{x}{2\left(x+2y\right)}\right)\)
\(\Rightarrow A\ge3-\frac{3}{2}+\frac{1}{2}\left(\frac{y}{y+2z}+\frac{z}{z+2x}+\frac{x}{x+2y}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}+\frac{x^2}{x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{y^2+2yz+z^2+2xz+x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}\right)\)
\(\Rightarrow A\ge2\)
Dấu " = " xảy ra khi x=y=z=1