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Theo đề bài thì ta có:
\(\frac{1}{a+b+1}=1-\frac{1}{b+c+1}+1-\frac{1}{c+a+1}=\frac{b+c}{b+c+1}+\frac{c+a}{c+a+1}\)
\(\ge2.\sqrt{\frac{\left(b+c\right)\left(c+a\right)}{\left(b+c+1\right)\left(c+a+1\right)}}\left(1\right)\)
Tương tự ta có:
\(\hept{\begin{cases}\frac{1}{b+c+1}\ge2.\sqrt{\frac{\left(a+b\right)\left(c+a\right)}{\left(a+b+1\right)\left(c+a+1\right)}1}\left(2\right)\\\frac{1}{c+a+1}\ge2.\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\left(3\right)\end{cases}}\)
Nhân (1), (2), (3) vế theo vế ta được
\(\frac{1}{a+b+1}.\frac{1}{b+c+1}.\frac{1}{c+a+1}\ge8.\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\frac{1}{8}\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{4}\)
Mình có cách này,không chắc lắm:
\(VT=\frac{a}{a\left(a^2+bc+1\right)}+\frac{b}{b\left(b^2+ac+1\right)}+\frac{c}{c\left(c^2+ab+1\right)}\) (làm tắt,bạn tự hiểu nha)
\(=\frac{1}{a^2+bc+1}+\frac{1}{b^2+ac+1}+\frac{1}{c^2+ab+1}\)
\(\le\frac{1}{3}\left(\frac{1}{\sqrt[3]{a}}+\frac{1}{\sqrt[3]{b}}+\frac{1}{\sqrt[3]{c}}\right)\)
\(=\frac{1}{3}\left[\left(1+1+1\right)-\left(\frac{\sqrt[3]{a}-1}{\sqrt[3]{a}}+\frac{\sqrt[3]{b}-1}{\sqrt[3]{b}}+\frac{\sqrt[3]{c}-1}{\sqrt[3]{c}}\right)\right]\)
\(=1-\frac{1}{3}\left(\frac{\sqrt[3]{a}-1}{\sqrt[3]{a}}+\frac{\sqrt[3]{b}-1}{\sqrt[3]{b}}+\frac{\sqrt[3]{c}-1}{\sqrt[3]{c}}\right)\)
Áp dụng BĐT Cô si với biểu thức trong ngoặc:
\(=1-\frac{1}{3}\left(\frac{\sqrt[3]{a}-1}{\sqrt[3]{a}}+\frac{\sqrt[3]{b}-1}{\sqrt[3]{b}}+\frac{\sqrt[3]{c}-1}{\sqrt[3]{c}}\right)\)
\(\le1-\sqrt[3]{\left(\sqrt[3]{a}-1\right)\left(\sqrt[3]{b}-1\right)\left(\sqrt[3]{c-1}\right)}\le1^{\left(đpcm\right)}\)
Dấu "=" xảy ra khi a = b = c = 1
Ta c/m bđt sau:
\(a^3+1\ge a^2+a\)
\(\Leftrightarrow a^3+1-a^2-a\ge0\Leftrightarrow a\left(a^2-1\right)-\left(a^2-1\right)\ge0\Leftrightarrow\left(a-1\right)^2\left(a+1\right)\ge0\)
\(\Rightarrow\frac{a}{a^3+a+1}\le\frac{a}{a^2+2a}=\frac{1}{a+2}\)
\(\Rightarrow\frac{a}{a^3+a+1}+\frac{b}{b^3+b+1}+\frac{c}{c^3+c+1}\le\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\)
Đặt \((a,b,c)\rightarrow(\frac{x}{y},\frac{y}{z},\frac{z}{x})\)
\(\Rightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=\frac{y}{x+2y}+\frac{z}{y+2z}+\frac{x}{z+2x}=\frac{1}{2}\left(1-\frac{x}{x+2y}+1-\frac{y}{y+2z}+1-\frac{z}{z+2x}\right)=\frac{3}{2}-\frac{1}{2}\left(\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xy}\right)\)\(\le\frac{3}{2}-\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}\right)=\frac{3}{2}-\frac{1}{2}.\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)
Dấu bằng xảy ra khi a=b=c=1
Ta có : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Leftrightarrow\left(3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)< 10\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{a}+\frac{a}{c}< 7\)
\(\Leftrightarrow\frac{a+c}{b}+\frac{b+a}{c}+\frac{c+b}{a}< 7\)
Không giảm tổng quá .Giả sử a là cạnh lớn nhất .Giả b + c < a => 0 < \(\frac{b+c}{a}\)
\(\Rightarrow\frac{a+c}{b}+\frac{b+a}{c}+\frac{c+b}{a}>\frac{2c+b}{b}+\frac{2b+c}{c}+\frac{b+c}{a}\)( không chắc lắm )
= \(\frac{2c}{b}+\frac{2b}{c}+\frac{b+c}{a}+2\)
=\(\frac{2\left(b+c\right)^2}{bc}+\frac{b+c}{a}-2>7\left(VL\right)\)
=>b+ c > a => a ; b ; c là 3 cạnh tam giác ( đpcm )
