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a+b=2=> a=2-b
\(\Rightarrow\left(1-\frac{4}{a^2}\right)\left(1-\frac{4}{b^2}\right)=\left(\frac{a^2-4}{a^2}\right)\left(\frac{b^2-4}{b^2}\right)=\frac{\left(2-b\right)^2-4}{\left(2-b\right)^2}.\frac{b^2-4}{b^2}\)
=\(\frac{b^2-2b-8}{b^2-2b}\)
đặt A=\(\frac{b^2-2b-8}{b^2-2b}\)
đkxđ \(\hept{\begin{cases}b\ne0\\b\ne2\end{cases}}\)
\(\Leftrightarrow Ab^2-2bA=b^2-2b-8\)
\(\Leftrightarrow\left(A-1\right)b^2-2\left(A-1\right)b+8=0\)
nếu A=1 => 8=0 (vô lý)
nếu A khác 1 pt có nghiệm khi \(\Delta\ge0\Leftrightarrow\left[-2\left(A-1\right)\right]^2-4\left(A-1\right).8\ge0\)
\(4A^2-40A+36\ge0\Leftrightarrow A^2-10A+9\ge0\Leftrightarrow\hept{\begin{cases}A\le1\\A\ge9\end{cases}}\)
GTNN A=9 dấu "=" <=> a=b=1
bạn ơi mình đặt nhầm B thành A rồi bn tự sửa lại nhé!
\(B=\left(1-\frac{4}{a^2}\right)\left(1-\frac{4}{b^2}\right)=\left(1-\frac{2}{a}\right)\left(1-\frac{2}{b}\right)\left(1+\frac{2}{a}\right)\left(1+\frac{2}{b}\right)\)
\(=\frac{\left(2-a\right)\left(2-b\right)\left(a+2\right)\left(b+2\right)}{a^2b^2}=\frac{ab.\left(a+2\right)\left(b+2\right)}{a^2b^2}=\frac{ab+2\left(a+b\right)+4}{ab}=\frac{8}{ab}+1\)
Theo BĐT Cauchy thì : \(a+b\ge2\sqrt{ab}\Rightarrow ab\le\frac{\left(a+b\right)^2}{4}\)
Suy ra : \(A\ge\frac{8}{\frac{2^2}{4}}+1=9\).Đẳng thức xảy ra khi a = b = 1/2
Vậy ......................................
Ta có
\(M=\left(1+a\right)\left(1+\frac{1}{b}\right)+\left(1+b\right)\left(1+\frac{1}{a}\right)=2+\frac{a}{b}+\frac{b}{a}+a+b+\frac{1}{a}+\frac{1}{b}\)
\(\ge2+2+a+b+\frac{4}{a+b}\)
\(=4+a+b+\frac{2}{a+b}+\frac{2}{a+b}\)
\(\ge4+2\sqrt{\left(a+b\right).\frac{2}{\left(a+b\right)}}+\frac{2}{\sqrt{2\left(a^2+b^2\right)}}\)
\(=4+2\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
\(A=\left(a+b+1\right)\left(a^2+b^2\right)+\frac{4}{a+b}+1-1\ge\left(a+b+1\right)2\sqrt{\left(ab\right)^2}+\frac{\left(2+1\right)^2}{a+b+1}-1\)
\(=2\left(a+b+1\right)+\frac{9}{a+b+1}-1\ge2\sqrt{ab}+1+2\sqrt{\frac{9\left(a+b+1\right)}{a+b+1}}-1\ge2+6=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}a^2=b^2\left(1\right)\\\frac{2}{a+b}=1\left(2\right)\\a+b+1=\frac{9}{a+b+1}\left(3\right)\end{cases}}\)
pt \(\left(1\right)\)\(\Leftrightarrow\)\(a=b\) ( vì a, b > 0 )
pt \(\left(2\right)\)\(\Leftrightarrow\)\(a=b=1\)
pt \(\left(3\right)\)\(\Leftrightarrow\)\(\left(a+b+1\right)^2=9\)\(\Leftrightarrow\)\(a+b+1=3\) ( đúng vì \(a=b=1\) )
Vậy GTNN của \(A\) là \(8\) khi \(a=b=1\)
Chúc bạn học tốt ~
Ồ sorry bạn nhiều, chỗ đấy bị lỗi kĩ thuật rồi, mình sửa lại nhé :
\(M\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
Lại có : \(\frac{ab+bc+ca}{2}\ge\frac{3\sqrt{a^3b^3c^3}}{2}=\frac{3}{2}\)
Do đó : \(M\ge\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Ta có : \(\frac{1}{a^3\left(b+c\right)}=\frac{\frac{1}{a^2}}{a\left(b+c\right)}=\frac{\left(\frac{1}{a}\right)^2}{a\left(b+c\right)}\)
Tương tự : \(\frac{1}{b^3\left(a+c\right)}=\frac{\left(\frac{1}{b}\right)^2}{b\left(a+c\right)}\) , \(\frac{1}{c^3\left(a+b\right)}=\frac{\left(\frac{1}{c}\right)^2}{c\left(a+b\right)}\)
Ta thấy : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Áp dụng BĐT Svacxo ta có :
\(M=\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^2\left(a+c\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\) \(\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Vâỵ \(M_{min}=\frac{3}{2}\) tại \(a=b=c=1\)
\(M=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)+x^3+\frac{1}{x^3}}\)
\(M=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\frac{2x^6+3x^4+3x^2+2}{x^3}}\)
\(M=\frac{\left[\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2\right]x^3}{2x^6+3x^4+3x^2+2}\)
\(M=\frac{x^3\left(6x^4+15x^2+\frac{15}{x^2}+\frac{6}{x^4}+18\right)}{2x^6+3x^4+3x^2+2}\)
\(M=\frac{\frac{6x^8+15x^6+18x^4+15x^2+6}{x^4}.x^3}{2x^6+3x^4+3x^2+2}\)
\(M=\frac{\frac{6x^8+15x^6+18x^4+15x^2+6}{x}}{2x^6+3x^4+3x^2+2}\)
\(M=\frac{6x^8+15x^6+18x^4+15x^2+6}{x\left(2x^6+3x^4+3x^2+2\right)}\)
\(M=\frac{3\left(x^2+1\right)^2\left(2x^4+x^2+2\right)}{x\left(x^2+1\right)\left(2x^4+x^2+2\right)}\)
\(M=\frac{3\left(x^3+1\right)}{x}\)
M=\(\frac{a^4}{a\left(b+1\right)^2}+\frac{b^4}{b\left(a+1\right)^2}\)
áp dụng bdt bunhiacopxki ta co
(a+b)M>=\(\left(\frac{a^2}{b+1}+\frac{b^2}{a+1}\right)^2\)
\(\left(\frac{a^2}{b+1}+\frac{b^2}{a+1}\right)^2>=\left[\frac{\left(a+b^2\right)}{a+1+b+1}\right]^2\)
\(=\frac{\left(a+b\right)^4}{\left(a+b+2\right)^2}>=\frac{\left(a+b\right)^4}{4\left(a+b\right)^2}\)(do 2<=a+b)
=\(\frac{\left(a+b\right)^2}{4}\)
do do M(a+b)>=\(\frac{\left(a+b\right)^2}{4}\)
=>M>=\(\frac{a+b}{4}>=\frac{1}{2}\)
dau = xay ra <=> a=b=1