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Do a,b,c là 3 cạnh tam giác nên \(a+b-c>0;b+c-a>0;c+a-b>0\)
Đặt \(x=b+c-a>0\)
\(y=a+c-b>0\)
\(z=a+b-c>0\)
\(\Rightarrow a=\frac{"y+z"}{2}\)
\(\Rightarrow b=\frac{"x+z"}{2}\)
\(\Rightarrow c=\frac{"x+y"}{2}\)
\(A=\frac{a}{"b+c-a"}+\frac{b}{"a+c-b"}+\frac{c}{"a+b-c"}\)
\(=\frac{"y+z"}{"2x"}+\frac{"x+z"}{"2y"}+\frac{"x+y"}{"2z"}\)
\(=\frac{1}{2}."\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}"\)
Áp dụng công thức bdt Cauchy cho 2 số :
\(\frac{x}{y}+\frac{y}{x}\ge2\)
\(\frac{x}{z}+\frac{z}{x}\ge2\)
\(\frac{y}{z}+\frac{z}{y}\ge2\)
Cộng 3 bdt trên, suy ra :
\("\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}"\ge6\)
\(\Rightarrow A\ge\frac{1}{2}.6=3\) "dpcm"
P/s: Nhớ thay thế dấu ngoặc kép thành dấu ngoặc đơn nhé
Lời giải:
Áp dụng BĐT AM-GM (Cô-si)
\(1+a^3+b^3\geq 3\sqrt[3]{a^3b^3}=3ab\)
\(\Rightarrow \frac{\sqrt{1+a^3+b^3}}{ab}\geq \frac{\sqrt{3ab}}{ab}=\frac{c\sqrt{3ab}}{abc}=c\sqrt{3ab}=\sqrt{c}.\sqrt{3abc}=\sqrt{3c}\)
Hoàn toàn tương tự:
\(\frac{\sqrt{1+b^3+c^3}}{bc}\geq \sqrt{3a}\)
\(\frac{\sqrt{1+a^3+c^3}}{ac}\geq \sqrt{3b}\)
Cộng theo vế những BĐT vừa thu được ta có:
\(\frac{\sqrt{a^3+b^3+1}}{ab}+\frac{\sqrt{b^3+c^3+1}}{bc}+\frac{\sqrt{c^3+a^3+1}}{ac}\geq \sqrt{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})\)
\(\geq \sqrt{3}.3\sqrt[3]{\sqrt{a}.\sqrt{b}.\sqrt{c}}=\sqrt{3}.3\sqrt[6]{abc}=3\sqrt{3}\) (áp dụng BĐT Cô-si)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
\(\sqrt{\dfrac{a}{b+c-ta}}=\dfrac{a\sqrt{t+1}}{\sqrt{\left(at+a\right)\left(b+c-ta\right)}}\ge\dfrac{2a\sqrt{t+1}}{at+a+b+c-ta}=\dfrac{2a\sqrt{t+1}}{a+b+c}\)
Làm tương tự, cộng lại và rút gọn
Áp dụng bđt cosi schwart ta có:
`VT>=(a+b+c)^2/(a+b+c+sqrt{ab}+sqrt{bc}+sqrt{ca})`
Dễ thấy `sqrt{ab}+sqrt{bc}+sqrt{ca}<a+b+c`
`=>VT>=(a+b+c)^2/(2(a+b+c))=(a+b+c)/2=3`
Dấu "=" `<=>a=b=c=1.`
\(T=\dfrac{a+b}{\sqrt{ab+c}}+\dfrac{b+c}{\sqrt{bc+a}}+\dfrac{c+a}{\sqrt{ca+b}}\)
\(\odot\) Ta có: \(\dfrac{a+b}{\sqrt{ab+c}}=\dfrac{a+b}{\sqrt{ab+c\left(a+b+c\right)}}=\dfrac{a+b}{\sqrt{\left(b+c\right)\left(a+c\right)}}\)
\(\odot\) Tương tự:
\(\dfrac{b+c}{\sqrt{bc+a}}=\dfrac{b+c}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\dfrac{c+a}{\sqrt{ca+b}}=\dfrac{c+a}{\sqrt{\left(a+b\right)\left(b+c\right)}}\)
\(\odot\) Áp dụng bất đẳng thức AM - GM
\(\Rightarrow T=\dfrac{a+b}{\sqrt{\left(a+c\right)\left(b+c\right)}}+\dfrac{b+c}{\sqrt{\left(a+c\right)\left(b+a\right)}}+\dfrac{a+c}{\sqrt{\left(a+b\right)\left(b+c\right)}}\)
\(\ge3\sqrt[3]{\dfrac{a+b}{\sqrt{\left(a+c\right)\left(b+c\right)}}\times\dfrac{b+c}{\sqrt{\left(a+c\right)\left(b+a\right)}}\times\dfrac{a+c}{\sqrt{\left(a+b\right)\left(b+c\right)}}}\)
\(=3\)
\(\odot\) Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(\left(b^3+c^3\right)\left(1+1\right)\left(1+1\right)\ge\left(b+c\right)^3\)
\(\Rightarrow b^3+c^3\ge\dfrac{\left(b+c\right)^3}{4}\Rightarrow\dfrac{a}{\sqrt[3]{b^3+c^3}}\le\dfrac{a\sqrt[3]{4}}{b+c}\)
Tương tự và cộng lại:
\(VT\le\sqrt[3]{4}\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)< \sqrt[3]{4}\left(\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}\right)=2\sqrt[3]{4}\)
set \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\)\(\Rightarrow x+y+z=3\)
\(VT=\sum\sqrt{\dfrac{\left(x+y\right)\left(x+z\right)}{4x}}=\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}.\left(\sum\dfrac{1}{\sqrt{4x\left(y+z\right)}}\right)\)
Áp dụng BĐT AM-GM:
\(\dfrac{1}{\sqrt{4x\left(y+z\right)}}+\dfrac{1}{\sqrt{4y\left(x+z\right)}}+\dfrac{1}{\sqrt{4z\left(x+y\right)}}\ge\dfrac{9}{2\left(\sqrt{xy+xz}+\sqrt{yz+yx}+\sqrt{xz+zy}\right)}\)
Áp dụng BĐT bunyakovsky:
\(\sum\sqrt{xy+yz}\le\sqrt{6\left(xy+yz+xz\right)}\)
\(\Rightarrow\sum\dfrac{1}{2\sqrt{x\left(y+z\right)}}\ge\dfrac{9}{2\sqrt{6\left(xy+yz+xz\right)}}\)
Mà \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)=\dfrac{8}{3}\left(xy+yz+xz\right)\)(*)
\(\Rightarrow VT\ge\sqrt{\dfrac{8}{3}\left(xy+yz+xz\right)}.\dfrac{9}{2\sqrt{6\left(xy+yz+xz\right)}}=3\)
Dấu = xảy ra khi x=y=z hay a=b=c=1
(*) Prove BĐT \(\left(m+n\right)\left(n+p\right)\left(m+p\right)\ge\dfrac{8}{9}\left(m+n+p\right)\left(mn+np+pm\right)\)
khai triển ,để ý rằng \(\left(m+n\right)\left(n+p\right)\left(p+m\right)=\left(m+n+p\right)\left(mn+np+pm\right)-mnp\)