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Ta có:
\(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge9\left(a+2b\right)\)
Mặt khác:
\(\left(a+2b\right)^2\le\left(1+2\right)\left(a^2+2b^2\right)\le3\times3c^2\)
\(\Rightarrow\left(a+2b\right)\le3c\)
\(\frac{9}{\left(a+2b\right)}\ge\frac{9}{3c}=\frac{3}{c}\)
\(=VT\ge\frac{3}{c}\left(ĐPCM\right)\)
Dấu "=" xảy ra khi a=b=c=1
![](https://rs.olm.vn/images/avt/0.png?1311)
http://diendantoanhoc.net/topic/118669-violympic9-c%C3%A1c-b%C3%A0i-to%C3%A1n-violympic-l%E1%BB%9Bp-9-cho-k%C3%AC-thi-qu%E1%BB%91c-gia-s%E1%BA%AFp-t%E1%BB%9Bi/page-5
http://diendantoanhoc.net/topic/136445-boxed-textbftopic-%C3%B4n-thi-violympic-qu%E1%BB%91c-gia-2014-2015/page-3
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Ta có \(\frac{a}{a^2+2b+3}=\frac{a}{a^2+1+2\left(b+1\right)}\le\frac{a}{2a+2\left(b+1\right)}=\frac{a}{2\left(a+b+1\right)}\)
Chứng minh tương tự \(\hept{\begin{cases}\frac{b}{b^2+2c+3}\le\frac{b}{2\left(b+c+1\right)}\\\frac{c}{c^2+2a+3}\le\frac{c}{2\left(a+c+1\right)}\end{cases}}\)
Cộng 3 vế của 3 bđt lại ta được
\(VT\le\frac{1}{2}\left(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\right)\)
Để bài toán được chứng minh thì ta cần \(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\le1\)
\(\Leftrightarrow1-\frac{a}{a+b+1}+1-\frac{b}{b+c+1}+1-\frac{c}{c+a+1}\ge2\)
\(\Leftrightarrow A=\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\ge2\)
Ta có \(A=\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\)
\(=\frac{\left(b+1\right)^2}{\left(b+1\right)\left(a+b+1\right)}+\frac{\left(c+1\right)^2}{\left(c+1\right)\left(b+c+1\right)}+\frac{\left(a+1\right)^2}{\left(a+1\right)\left(c+a+1\right)}\)
Áp dụng bđt quen thuộc \(\frac{m^2}{x}+\frac{n^2}{y}+\frac{p^2}{z}\ge\frac{\left(m+n+p\right)^2}{x+y+z}\)(quen thuộc) ta được
\(A\ge\frac{\left(a+b+c+3\right)^2}{\left(b+1\right)\left(a+b+1\right)+\left(c+1\right)\left(b+c+1\right)+\left(a+1\right)\left(c+a+1\right)}\)
\(=\frac{\left(a+b+c+3\right)^2}{a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3}\)
\(=\frac{2\left(a+b+c+3\right)^2}{2\left(a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3\right)}\)
\(=\frac{2\left(a+b+c+3\right)^2}{a^2+b^2+c^2+\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)+6\left(a+b+c\right)+6}\)
\(=\frac{2\left(a+b+c+3\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)+6\left(a+b+c\right)+9}\)
\(=\frac{2\left(a+b+c+3\right)^2}{\left(a+b+c+3\right)^2}=2\)(DDpcm)
Dấu "=" xảy ra tại a= b = c =1
bn có thể ghi cho mk cái bđt đấy đc ko
#mã mã#
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\(\frac{a}{b^2+c^2}=\frac{a}{1-a^2}=\frac{a^2}{a-a^3}\)
Chứng minh: \(a-a^3\le\frac{2}{3\sqrt{3}}\text{ }\left(#\right)\)
\(\left(#\right)\Leftrightarrow a^3-a+\frac{2}{3\sqrt{3}}\ge0\Leftrightarrow\left(a-\frac{1}{\sqrt{3}}\right)^2\left(x+\frac{2}{\sqrt{3}}\right)\ge0\)
Bất đẳng thức cuối đúng nên có đpcm.
\(\Rightarrow P\ge\frac{1}{\frac{2}{3\sqrt{3}}}\left(a^2+b^2+c^2\right)=\frac{3\sqrt{3}}{2}\)