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Vì a và 1 là 2 số dương \(\Rightarrow a+1\ge2\sqrt{a}\) (bđt AM - GM)
Vì b và 1 là 2 số dương \(\Rightarrow b+1\ge2\sqrt{b}\)(bđt AM - GM)
Vì c và 1 là 2 số dương \(\Rightarrow c+1\ge2\sqrt{c}\)(bđt AM - GM)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge2\sqrt{a}.2\sqrt{b}.2\sqrt{c}=8\sqrt{abc}=8\)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\) (đpcm)
Nhân tung tóe + rút gọn ta được: \(\Sigma_{cyc}a^3b^2+\Sigma_{cyc}ab^3\ge abc\left(ab+bc+ca+a+b+c\right)\)
\(\Leftrightarrow\)\(\Sigma\frac{a^2b}{c}+\Sigma\frac{a^2}{b}\ge ab+bc+ca+a+b+c\) (*)
(*) đúng do \(\hept{\begin{cases}\frac{a^2b}{c}+bc\ge2ab\\\frac{a^2}{b}+b\ge2a\end{cases}}\Rightarrow\hept{\begin{cases}\Sigma\frac{a^2b}{c}\ge ab+bc+ca\\\Sigma\frac{a^2}{b}\ge a+b+c\end{cases}}\)
"=" \(\Leftrightarrow\)\(a=b=c\)
Ta có \(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)\(\Rightarrow3\sqrt[3]{a^2b^2c^2}\le3\Leftrightarrow abc\le1\)
\(\Rightarrow\)\(\frac{1}{1+a^2\left(b+c\right)}\le\frac{1}{abc+a^2\left(b+c\right)}\)\(=\frac{1}{a\left(ab+bc+ca\right)}=\frac{1}{3a}\)
\(CMTT\Rightarrow\frac{1}{1+b^2\left(c+a\right)}\le\frac{1}{3b}\)
\(\frac{1}{1+c^2\left(a+b\right)}\le\frac{1}{3c}\)
\(\Rightarrow VT\le\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}\)\(=\frac{ab+bc+ca}{3abc}=\frac{1}{abc}\)
\(1+a^2=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)
Tương tự, ta có: \(1+b^2=\left(a+b\right)\left(b+c\right)\)\(;\)\(1+c^2=\left(b+c\right)\left(c+a\right)\)
\(\Rightarrow\)\(\frac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) ( do a, b, c dương )
\(\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}=\frac{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
...
Ta có:\(\left(a-1\right)^2\ge0\)
\(\Leftrightarrow a^2-2a+1\ge0\)
\(\Leftrightarrow\left(a^2+2a+1\right)-4a\ge0\)
\(\Leftrightarrow\left(a+1\right)^2\ge4a\)
TT\(\Rightarrow\left(b+1\right)^2\ge4b\)
\(\left(c+1\right)^2\ge4b\)
Nhân vế theo vế ta được \(\left[\left(a+1\right)\left(b+1\right)\left(c+1\right)\right]^2\ge64abc=64\)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\)(đpcm)
Chứng minh BĐT Phụ: \(a^5+b^5\ge a^4b+ab^4\)với \(a;b>0\)
\(\Rightarrow\frac{a^5+b^5}{ab\left(a+b\right)}\ge\frac{a^4b+ab^4}{ab\left(a+b\right)}=\frac{ab\left(a^3+b^3\right)}{ab\left(a+b\right)}=\frac{ab\left(a+b\right)\left(a^2-ab+b^2\right)}{ab\left(a+b\right)}=a^2-ab+b^2\)
Áp dụng ta có: \(VT\)(VẾ TRÁI)\(\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\) \(\left(1\right)\)
Xét: \(\left[2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\right]-\left[3\left(ab+bc+ca\right)-2\right]\)
\(=2\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)+2\)
\(=4\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)\) (Do a2+b2+c2=1) \(\left(2\right)\)
Mà \(a^2+b^2+c^2\ge ab+bc+ca\) Tự chứng minh \(\left(3\right)\)
Từ (1);(2) và (3) suy ra \(VT\ge3\left(ab+bc+ca\right)-2\)
Vậy \(\frac{a^5+b^5}{ab\left(a+b\right)}+\frac{b^5+c^5}{bc\left(b+c\right)}+\frac{c^5+a^5}{ca\left(c+a\right)}\ge3\left(ab+bc+ca\right)-2\)
\(\Leftrightarrow\left(a+b\right)\left(ab+abc\right)\ge4ab\)
\(\Leftrightarrow\left(a+b\right)\left(c+1\right)\ge4\)(1)
Lại có:\(\left(a+b\right)^2\ge4ab\)
\(\left(c+1\right)^2\ge4c\)
\(\Rightarrow\left[\left(a+b\right)\left(c+1\right)\right]^2\ge16abc=16\)
\(\Rightarrow\left(a+b\right)\left(c+1\right)\ge4\)
\(\Rightarrow\left(1\right)\) đúng\(\Rightarrowđpcm\)