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\(a+b+c=7\Rightarrow a+b+c-1=6\)
Ta có:\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow49=23+2\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca=13\)
Lại có \(ab+c-6=ab+c-\left(a+b+c-1\right)=ab-a-b+1=\left(a-1\right)\left(b-1\right)\)
Tương tự \(bc+a-6=\left(b-1\right)\left(c-1\right)\)
\(ca+b-6=\left(c-1\right)\left(a-1\right)\)
\(\Rightarrow A=\frac{1}{\left(a-1\right)\left(b-1\right)}+\frac{1}{\left(b-1\right)\left(c-1\right)}+\frac{1}{\left(c-1\right)\left(a-1\right)}\)
\(=\frac{c-1+a-1+b-1}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=\frac{a+b+c-3}{abc-\left(ab+ac+bc\right)+\left(a+b+c\right)-1}\)
\(=\frac{7-3}{3-13+7-1}=-1\)
Vì a,b>0 nên:\(ab>0;\left(a^2-b^2\right)^2\ge0\)
\(\Leftrightarrow ab\left(a^2-b^2\right)^2\ge0\)
\(\Leftrightarrow ab\left(a^4-2a^2b^2+b^4\right)\ge0\)
\(\Leftrightarrow a^5b-2a^3b^3+ab^5\ge0\)
\(\Leftrightarrow a^6+ab^5+a^5b+b^6-a^6-2a^3b^3-b^6\ge0\)
\(\Leftrightarrow a\left(a^5+b^5\right)+b\left(a^5+b^5\right)-\left(a^3+b^3\right)^2\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a^5+b^5\right)\ge\left(a^3+b^3\right)^2\)
\(\Leftrightarrow a+b\ge a^3+b^3\)(Vì a^5+b^5=a^3+b^3 và a^3+b^3;a^5+b^5>0)
\(\Leftrightarrow a+b\ge\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\Leftrightarrow a^2-ab+b^2\ge1\)
Vậy GTLN M=1 tại \(a^2-b^2=0\Leftrightarrow a=b\)
\(\Leftrightarrow a^3+a^3=a^5+a^5\)(Vì a=b)
\(\Leftrightarrow\orbr{\begin{cases}a=0\\a=1\end{cases}}\)(TH a=0 loại vì a>0)
\(\Leftrightarrow b=1\)
\(a^2+b^2\le1+ab\)
\(\Leftrightarrow a^2-ab+b^2\le1\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\le a+b\)
\(\Leftrightarrow a^3+b^3\le a+b\)
\(\Leftrightarrow\left(a^3+b^3\right)\left(a^3+b^3\right)\le\left(a+b\right)\left(a^5+b^5\right)\) ( \(a^3+b^3=a^5+b^5\))
\(\Leftrightarrow a^6+2a^3b^3+b^6\le a^6+ab^5+a^5b+b^6\)
\(\Leftrightarrow a^5b+ab^5\ge2a^3b^3\)
\(\Leftrightarrow a^5b+ab^5-2a^3b^3\ge0\)
\(\Leftrightarrow ab\left(a^4-2a^2b^2+b^4\right)\ge0\)
\(\Leftrightarrow ab\left(a^2-b^2\right)^2\ge0\) (luôn đúng \(\forall a;b>0\))
Vậy \(a^2+b^2\le1+ab\)
Đặt ab + bc + ca = q; abc = r. Ta có:
\(A=\dfrac{\left(ab+bc+ca\right)+6\left(a+b+c\right)+27}{abc+3\left(ab+bc+ca\right)+9\left(a+b+c\right)+27}-\dfrac{1}{3\left(ab+bc+ca\right)}\)
\(A=\dfrac{q+33}{r+3q+36}-\dfrac{1}{3q}\).
Theo bất đẳng thức Schur: \(a^3+b^3+c^3+3abc\ge a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\)
\(\Leftrightarrow\left(a+b+c\right)^3+9abc\ge4\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\Leftrightarrow9r\ge4q-1\Leftrightarrow r\ge\dfrac{4q-1}{9}\).
Từ đó \(A\le\dfrac{q+33}{\dfrac{4q-1}{9}+3q+36}-\dfrac{1}{3q}\)
\(\Rightarrow A\leq \frac{27q^2+860q-323}{93q^2+969q}\)
\(\Rightarrow A+\dfrac{1}{10}=\dfrac{\left(3q-1\right)\left(121q+3230\right)}{30q\left(31q+323\right)}\le0\). (Do \(q=ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\))
\(\Rightarrow A\leq \frac{-1}{10}\). Dấu "=" xảy ra khi và chỉ khi a = b = c = 1.
\(a^3+b^3=a^5+b^5\)
\(\Rightarrow\left(a^2.a^1\right)+\left(b^2.b^1\right)=\left(a^2.a^3\right)+\left(b^2.b^3\right)\)
\(\Rightarrow\dfrac{a^2}{a^2}=\dfrac{a^3+\left(b^2+b^3\right)}{a^1+b^2.b^1}\)
\(\Rightarrow1=a^2+b^2\)
\(\Rightarrow1\ge\left(a^2+b^2\right)-ab\)
\(\Rightarrow1+ab\le a^2+b^2\)