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Ta có: \(2x^2+\frac{y^2}{4}+\frac{1}{x^2}=4\)
=> \(\left(x^2+\frac{y^2}{4}\right)+\left(x^2+\frac{1}{x^2}\right)=4\)
Lại có: \(x^2+\frac{y^2}{4}\ge2.x.\frac{y}{2}=xy\) Và \(x^2+\frac{1}{x^2}\ge2.x.\frac{1}{x}=2\)
=> \(4\ge xy+2\)=> \(2\ge xy\)
=> \(A=2016+xy\le2016+2=2018\)
=> Amin=2018
\(\sqrt[]{\sqrt{ }\frac{ }{ }\sqrt[]{}3\hept{\begin{cases}\\\\\end{cases}}3\frac{ }{ }\sqrt{ }\cos\hept{\begin{cases}\\\\\end{cases}}\Omega3\cong}\)
b) \(\left(1+a\right).\frac{1}{1+b^2}=\left(1+a\right)\left(1-\frac{b^2}{1+b^2}\right)\)
\(\ge\left(1+a\right)\left(1-\frac{b^2}{2b}\right)=1+a-\frac{ab+b}{2}\)
Thiết lập hai BĐT còn lại tương tự và cộng theo vế được:
\(VT\ge6-\frac{ab+bc+ca+3}{2}\ge6-\frac{\frac{\left(a+b+c\right)^2}{3}+3}{2}\)
\(=6-\frac{3+3}{2}=3^{\left(đpcm\right)}\)
Dấu "=" xảy ra khi a = b = c = 1
1) \(21x^2+21y^2+z^2\)
\(=18\left(x^2+y^2\right)+z^2+3\left(x^2+y^2\right)\)
\(\ge9\left(x+y\right)^2+z^2+3.2xy\)
\(\ge2.3\left(x+y\right).z+6xy\)
\(=6\left(xy+yz+zx\right)=6.13=78\)
Dấu "=" xảy ra <=> x = y ; 3(x+y) = z; xy + yz + zx= 13 <=> x = y = 1; z= 6
2) \(x+y+z=3xyz\)
<=> \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=3\)
Đặt: \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)=> ab + bc + ca = 3
Ta cần chứng minh: \(3a^2+b^2+3c^2\ge6\)
Ta có: \(3a^2+b^2+3c^2=\left(a^2+c^2\right)+2\left(a^2+c^2\right)+b^2\)
\(\ge2ac+\left(a+c\right)^2+b^2\ge2ac+2\left(a+c\right).b=2\left(ac+ab+bc\right)=6\)
Vậy: \(\frac{3}{x^2}+\frac{1}{y^2}+\frac{3}{z^2}\ge6\)
Dấu "=" xảy ra <=> a = c = \(\sqrt{\frac{3}{5}}\); \(b=2\sqrt{\frac{3}{5}}\)
khi đó: \(x=z=\sqrt{\frac{5}{3}};y=\sqrt{\frac{5}{3}}\)
1a
\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)
\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(A_{min}=\frac{161}{16}\)
1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)
\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)