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\(A=\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}\ge\dfrac{4}{2b}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{4}{b+c-a+c+a-b}\ge\dfrac{4}{2c}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{4}{a+b-c+c+a-b}\ge\dfrac{4}{2a}\ge\dfrac{2}{a}\end{matrix}\right.\)
\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow A\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(dấu"="xảy\) \(ra\Leftrightarrow a=b=c\)
Đặt b + c - a = x; c + a - b = y; a + b - c = z. (x, y, z > 0)
Ta có \(A=\dfrac{a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{9c}{a+b-c}=\dfrac{y+z}{2x}+\dfrac{2\left(z+x\right)}{y}+\dfrac{9\left(x+y\right)}{2z}=\left(\dfrac{y}{2x}+\dfrac{2x}{y}\right)+\left(\dfrac{z}{2x}+\dfrac{9x}{2z}\right)+\left(\dfrac{9y}{2z}+\dfrac{2z}{y}\right)\ge2\sqrt{\dfrac{y}{2x}.\dfrac{2x}{y}}+2\sqrt{\dfrac{z}{2x}.\dfrac{9x}{2z}}+2\sqrt{\dfrac{9y}{2z}.\dfrac{2z}{y}}=2+3+6=11\).
Dấu "=" xảy ra khi và chỉ khi \(3y=2z=6x\Leftrightarrow3\left(c+a-b\right)=2\left(b+c-a\right)=6\left(a+b-c\right)\)
\(\Leftrightarrow a=\dfrac{5}{6};b=\dfrac{2}{3};c=\dfrac{1}{2}\).
\(\sqrt{\dfrac{a}{b+c-ta}}=\dfrac{a\sqrt{t+1}}{\sqrt{\left(at+a\right)\left(b+c-ta\right)}}\ge\dfrac{2a\sqrt{t+1}}{at+a+b+c-ta}=\dfrac{2a\sqrt{t+1}}{a+b+c}\)
Làm tương tự, cộng lại và rút gọn
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 \(\left(m+n\right)\left(n+p\right)\left(p+m\right)=\left(m+n+p\right)\left(mn+np+pm\right)-mnp\)
BĐT\(\Leftrightarrow\dfrac{a}{-a+b+c}+\dfrac{b}{a-b+c}+\dfrac{c}{a+b-c}\ge3\)
Áp dụng BĐT Svac-xơ, ta có:
\(\dfrac{a^2}{-a^2+ab+ac}+\dfrac{b^2}{ab-b^2+bc}+\dfrac{c^2}{ac+bc-c^2}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)}\)
Ta có: \(a,b,c\) là 3 cạnh của 1 tam giác nên:
\(a\left(b+c\right)>a^2\). Tương tự và cộng theo vế, ta có:
\(2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)>0\)
Ta sẽ chứng minh \(\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)}\ge3\left(1\right)\)
Thật vậy, \(BĐT\left(1\right)\Leftrightarrow3\left(a^2+b^2+c^2\right)+\left(a+b+c\right)^2\ge6\left(ab+bc+ca\right)\), đúng
Đẳng thức xảy ra khi và chỉ khi \(a=b=c\)
Cách 2:
Đặt \(\left\{{}\begin{matrix}-a+b+c=x\\a-b+c=y\\a+b-c=z\end{matrix}\right.\) với \(x,y,z>0\)
Khi đó ta có \(a=\dfrac{y+z}{2};b=\dfrac{x+z}{2};c=\dfrac{x+y}{2}\)
BĐT cần chứng minh trở thành:
\(\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\ge6\), đúng theo bđt Cauchy
Đẳng thức xảy ra khi và chỉ khi \(x=y=z\Leftrightarrow a=b=c\)