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\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
ta có :\(a^2-ab+b^2=\left(a+b\right)^2-3ab\ge\left(a+b\right)^2-\dfrac{3}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)(theo BĐT AM-GM)
\(\Rightarrow P\ge\sum\dfrac{a+b}{2\sqrt{ab+1}}\)
ÁP dụng BĐT AM-GM:
\(\dfrac{a+b}{2\sqrt{ab+1}}+\dfrac{b+c}{2\sqrt{bc+1}}+\dfrac{c+a}{2\sqrt{ca+1}}\ge3\sqrt[3]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}=\dfrac{3}{2}.\dfrac{1}{\sqrt[3]{\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}\)
Mà \(\sqrt[3]{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\le\dfrac{1}{3}\left(ab+bc+ca+3\right)\)
\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\left(ab+bc+ca+3\right)}}\)(*)
ta liên tưởng đến BĐT phụ:\(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)
Cm: phân tích :\(VT=xy\left(x+y\right)+yz\left(y+z\right)+zx\left(x+z\right)+2xyz\)
\(=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(z+x\right)+3xyz-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)
mà \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}=9xyz\)
nên \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+xz\right)=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
Áp dụng:
\(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
mặt khác,theo AM-GM,dễ dàng chứng minh được \(a+b+c\ge\dfrac{3}{2}\)
nên \(1\ge\dfrac{8}{9}.\dfrac{3}{2}\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le\dfrac{3}{4}\)
từ (*)\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\dfrac{3}{4}+3}}=\dfrac{3}{\sqrt{5}}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{2}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{\left(1-a\right)^2}+\dfrac{1-a}{8}+\dfrac{1-a}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{\left(1-b\right)^2}+\dfrac{1-b}{8}+\dfrac{1-b}{8}\ge\dfrac{3b}{4}\\\dfrac{c^3}{\left(1-c\right)^2}+\dfrac{1-c}{8}+\dfrac{1-c}{8}\ge\dfrac{3c}{4}\end{matrix}\right.\)
\(\Rightarrow P+\dfrac{6-2\left(a+b+c\right)}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow P\ge\dfrac{1}{4}\)
Vậy \(P_{min}=\dfrac{1}{4}\)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(P=\dfrac{9}{ab+bc+ca}+\dfrac{2}{a^2+b^2+c^2}\)
\(=2\left[\dfrac{1}{a^2+b^2+c^2}+\dfrac{4}{2\left(ab+bc+ca\right)}\right]+\dfrac{5}{ab+bc+ca}\)
\(\ge2.\dfrac{\left(1+2\right)^2}{\left(a+b+c\right)^2}+\dfrac{5}{ab+bc+ca}\)
\(=\dfrac{18}{1}+\dfrac{5}{ab+bc+ca}\ge18+5.\dfrac{3}{\left(a+b+c\right)^2}=18+15=33\)
Đẳng thức xảy ra khi a=b=c=1/3.
Vậy GTNN của P là 33.
Ta có \(ab+bc+ca=2abc\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
Đặt \(\left\{{}\begin{matrix}x=\dfrac{1}{a}\\y=\dfrac{1}{b}\\z=\dfrac{1}{c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=2\\P=\dfrac{x^3}{\left(2-x\right)^2}+\dfrac{y^3}{\left(2-y\right)^3}+\dfrac{z^3}{\left(2-z\right)^2}\end{matrix}\right.\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{x^3}{\left(2-x\right)^2}+\dfrac{2-x}{8}+\dfrac{2-x}{8}\ge3\sqrt[3]{\dfrac{x^3}{64}}=\dfrac{3x}{4}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^3}{\left(2-y\right)^2}+\dfrac{2-y}{8}+\dfrac{2-y}{8}\ge\dfrac{3y}{4}\\\dfrac{z^3}{\left(2-z\right)^2}+\dfrac{2-z}{8}+\dfrac{2-z}{8}\ge\dfrac{3z}{8}\end{matrix}\right.\)
\(\Rightarrow P+\dfrac{12-2\left(x+y+z\right)}{8}\ge\dfrac{3}{4}\left(x+y+z\right)\)
\(\Rightarrow P\ge\dfrac{1}{2}\)
Dấu " = " xảy ra khi \(x=y=z=\dfrac{2}{3}\)
Ta có : \(ab+bc+ca=2abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z\right)^2}\end{cases}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)
Tương tự ta có :
\(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)
\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)
\(\Rightarrow P\ge\frac{1}{12}\)
Dấu " = " xảy ra khi \(x=y=z=\frac{2}{3}\)
Dự đoán dấu "=" khi \(a=b=c \Rightarrow P=28\)
Ta sẽ chứng minh \(P=28\) là GTNN
Thật vậy ta có: \(P=\dfrac{ab+bc+ca}{a^2+b^2+c^2}-1+\dfrac{\left(a+b+c\right)^3}{abc}-27\ge0\)
\(\Leftrightarrow\dfrac{ab+bc+ca-\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}+\dfrac{\left(a+b+c\right)^3-27abc}{abc}\ge0\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^3-27abc}{abc}-\dfrac{2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)}{2\left(a^2+b^2+c^2\right)}\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(\dfrac{\dfrac{a+b+7c}{2}\cdot\left(a-b\right)^2}{abc}-\dfrac{\left(a-b\right)^2}{2\left(a^2+b^2+c^2\right)}\right)\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{a+b+7c}{2abc}-\dfrac{1}{2\left(a^2+b^2+c^2\right)}\right)\right)\ge0\) *Đúng*
Vậy ...
Đẳng cấp !!