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Hình như bạn viết nhầm đề, làm gì có số 9 ở đầu?
\(\frac{1}{1+a}+\frac{1}{1+b}\ge2\sqrt{\frac{1}{\left(1+a\right)\left(1+b\right)}}\)
\(\frac{a}{1+a}+\frac{b}{1+b}\ge2\sqrt{\frac{ab}{\left(1+a\right)\left(1+b\right)}}\)
Cộng vế với vế: \(1\ge\frac{1+\sqrt{ab}}{\sqrt{\left(1+a\right)\left(1+b\right)}}\Leftrightarrow\left(1+a\right)\left(1+b\right)\ge\left(1+\sqrt{ab}\right)^2\)
Áp dụng xuống dưới ta có:
\(M\ge\left(1+\sqrt{b}\right)^2\left(1+\frac{4}{\sqrt{b}}\right)^2=\left(5+\frac{4}{\sqrt{b}}+\sqrt{b}\right)^2\ge\left(5+2\sqrt{\frac{4\sqrt{b}}{\sqrt{b}}}\right)^2=81\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}b=4\\a=2\end{matrix}\right.\)
Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)
\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)
Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Ai có cách hay?
1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.
2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)
\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)
\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)
Áp dụng BĐT AM - GM
\(A=\left(a+1\right)\left(1+\frac{1}{b}\right)+\left(b+1\right)\left(1+\frac{1}{a}\right)\)
\(=\frac{a}{b}+\frac{b}{a}+a+\frac{1}{a}+b+\frac{1}{b}+2\)
\(=\frac{a}{b}+\frac{b}{a}+\left(a+\frac{1}{2a}\right)+\left(b+\frac{1}{2b}\right)+\frac{1}{2a}+\frac{1}{2b}+2\)
\(\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{a.\frac{1}{2a}}+2\sqrt{b.\frac{1}{2b}}+2\sqrt{\frac{1}{2a}.\frac{1}{2b}}+2\)
\(=4+2\sqrt{2}+\frac{1}{\sqrt{ab}}\ge4+2\sqrt{2}+\frac{1}{\frac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)
\(=4+3\sqrt{2}\)
Dấu " = " xảy ra khi \(a=b=\frac{1}{\sqrt{2}}\)
Ta co:\(1=a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\Rightarrow a+b\le\sqrt{2}\)
Ta lai co:
\(A=\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b}+a+b+2\)
\(=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{1}{a}+2a\right)+\left(\frac{1}{b}+2b\right)-\left(a+b\right)+2\)
\(\ge2+2\sqrt{2}+2\sqrt{2}-\sqrt{2}+2=4+3\sqrt{2}\)
Dau '=' xay ra khi \(a=b=\frac{1}{\sqrt{2}}\)
Vay \(A_{min}=4+3\sqrt{2}\)khi \(a=b=\frac{1}{\sqrt{2}}\)
Áp dụng BĐT AM - GM:
\(\frac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\) \(\Rightarrow abc\le\frac{1}{8}\)
\(1+1+1+\frac{1}{2a}+\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2b}\ge7\sqrt[7]{\frac{1}{16a^2b^2}}\)
\(\Leftrightarrow3+\frac{1}{a}+\frac{1}{b}\ge7\sqrt[7]{\frac{1}{16a^2b^2}}\)
Tương tự ta CM được:
\(3+\frac{1}{b}+\frac{1}{c}\ge7\sqrt[7]{\frac{1}{16b^2c^2}}\)
\(3+\frac{1}{c}+\frac{1}{a}\ge\ge7\sqrt[7]{\frac{1}{16c^2a^2}}\)
Nhân vế theo vế 3 bất đẳng thức trên:
\(S\ge343\sqrt[7]{\frac{1}{4096a^4b^4c^4}}\ge343\sqrt[7]{\frac{1}{4096.\frac{1}{8^4}}}=343\)
\(\Rightarrow Min_S=343\Leftrightarrow a=b=c=\frac{1}{2}\)
Tìm GTLN ko phải tìm GTNN
Ta có: \(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}=1\) (*)
Lại có: \(\left(a+1\right)^2+b^2+1=a^2+b^2+2a+2\ge2ab+2a+2=2\left(ab+a+1\right)\)
\(\Rightarrow\frac{1}{\left(a+1\right)^2+b^2+1}\le\frac{1}{2\left(ab+a+1\right)}\) tương tự ta có:
\(\frac{1}{\left(b+1\right)^2+c^2+1}\le\frac{1}{2\left(bc+b+1\right)};\frac{1}{\left(c+1\right)^2+a^2+1}\le\frac{1}{2\left(ca+c+1\right)}\)
Cộng theo vế ta có: \(P\le\frac{1}{2\left(ab+a+1\right)}+\frac{1}{2\left(bc+b+1\right)}+\frac{1}{2\left(ca+c+1\right)}\)
\(=\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)=\frac{1}{2}\) theo (*)
Dấu "=" khi a=b=c=1
\(P=\left[\left(2+\frac{1}{a}+\frac{1}{b}\right)+1\right]\left[\left(2+\frac{1}{b}+\frac{1}{c}\right)+1\right]\left[\left(2+\frac{1}{c}+\frac{1}{a}\right)+1\right]\)
\(\ge\left(6\sqrt[3]{\frac{1}{4ab}}+1\right)\left(6\sqrt[3]{\frac{1}{4bc}}+1\right)\left(6\sqrt[3]{\frac{1}{4ca}}+1\right)\)
\(\ge\left[7\sqrt[7]{\left(\sqrt[3]{\frac{1}{4ab}}\right)^6}\right]\left[7\sqrt[7]{\left(\sqrt[3]{\frac{1}{4bc}}\right)^6}\right]\left[7\sqrt[7]{\left(\sqrt[3]{\frac{1}{4ca}}\right)^6}\right]\)
\(=\left[7\sqrt[7]{\left(\frac{1}{4ab}\right)^2}\right]\left[7\sqrt[7]{\left(\frac{1}{4bc}\right)^2}\right]\left[7\sqrt[7]{\left(\frac{1}{4ca}\right)^2}\right]\)
\(=343\sqrt[7]{\left(\frac{1}{64\left(abc\right)^2}\right)^2}\ge343\sqrt[7]{\left(\frac{1}{64\left[\frac{\left(a+b+c\right)^3}{27}\right]^2}\right)^2}=343\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)
P/s: Em chưa check lại đâu nha::D
Khúc cuối bài ban nãy là \(\ge343\) nha! Em đánh nhầm
Cách khác (em thử dùng Holder, mới học nên em không chắc lắm):
\(P\ge\left(3+\sqrt[3]{\frac{1}{abc}}+\sqrt[3]{\frac{1}{abc}}\right)^3=\left(3+2\sqrt[3]{\frac{1}{abc}}\right)^3\ge\left(3+2\sqrt[3]{\frac{1}{\left[\frac{\left(a+b+c\right)^3}{27}\right]}}\right)^3\ge343\)