P(x)=\(ax^2+bx+c\) (1)(a\(\ne0\) )

Ta có : 

\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}\)\(\Rightarrow\left\{{}\begin{matrix}b=2a\\c=3a\end{matrix}\right.\)(2)

Thay(2) vào (1)\(\Rightarrow P\left(x\right)=ax^2+2ax+3a\)

\(\Rightarrow\dfrac{P\left(-2\right)-3P\left(-1\right)}{a}=\dfrac{4a-4a+3a-3\left(a-2a+3a\right)}{a}\)=\(\dfrac{3a-3a+6a-9a}{a}=\dfrac{-3a}{a}=-3\)

\(\Leftrightarrow a^2\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+b^2\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)+c^2\left(\dfrac{1}{c+a}-\dfrac{1}{a+b}\right)=0\)

\(\Leftrightarrow\dfrac{a^2\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{b^2\left(a-b\right)}{\left(a+c\right)\left(b+c\right)}+\dfrac{c^2\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}=0\)

\(\Leftrightarrow a^2\left(c-a\right)\left(c+a\right)+b^2\left(a-b\right)\left(a+b\right)+c^2\left(b-c\right)\left(b+c\right)=0\)

\(\Leftrightarrow a^2\left(c^2-a^2\right)+b^2\left(a^2-b^2\right)+c^2\left(b^2-c^2\right)=0\)

\(\Leftrightarrow a^2c^2+a^2b^2+b^2c^2-a^4-b^4-c^4=0\)

\(\Leftrightarrow2a^4+2b^4+2c^4-2a^2b^2-2a^2c^2-2b^2c^2=0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(a^2-c^2\right)^2+\left(b^2-c^2\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}a^2-b^2=0\\a^2-c^2=0\\b^2-c^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=0\\\left(a-c\right)\left(a+c\right)=0\\\left(b-c\right)\left(b+c\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\) (do \(\left(a+b\right)\left(a+c\right)\left(b+c\right)\ne0\) \(\Rightarrow\left\{{}\begin{matrix}a+b\ne0\\a+c\ne0\\b+c\ne0\end{matrix}\right.\))

\(\Rightarrow a=b=c\)

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\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)

Ta có \(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+bc+ca}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)

Tương tự: \(\left\{{}\begin{matrix}\dfrac{b^2}{b+ca}=\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}\\\dfrac{c^2}{c+ba}=\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\end{matrix}\right.\)

Áp dụng BĐT cosi:

\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3}{4}a\)

\(\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3}{4}b\)

\(\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{c^3}{64}}=\dfrac{3}{4}c\)

Cộng VTV:

\(\Leftrightarrow VT+\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\\ \Leftrightarrow VT\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{2\left(a+b+c\right)}{8}\\ \Leftrightarrow VT\ge\dfrac{a+b+c}{4}\)

Dấu \("="\Leftrightarrow a=b=c=3\)

Với cả 3 phần thì dấu "=" xảy ra tại a=b=c=1.

a) \(\dfrac{a}{1+b^2}=\dfrac{a\left(1+b^2\right)}{1+b^2}-\dfrac{ab^2}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\)

(Cosi) \(\ge a-\dfrac{ab^2}{2b}=a-\dfrac{ab}{2}\)

Tương tự : \(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2};\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\)

\(\Rightarrow P\ge\left(a+b+c\right)-\dfrac{ab+bc+ca}{2}\ge\left(CS\right)\left(a+b+c\right)-\dfrac{\left(a+b+c\right)^2}{6}=3-\dfrac{3^2}{6}=\dfrac{3}{2}\)

b) \(\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge\left(CS\right)1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}\)

Tương tự : \(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2};\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)

\(\Rightarrow P\ge3-\dfrac{a+b+c}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\)

c)\(P=\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}=\left(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\right)+\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\ge\dfrac{3}{2}+\dfrac{3}{2}=3\)