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Bất đẳng thức sai, chẳng hạn với \(a=b=10^{-4};c=0,5-a-b\).
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
Đề bài có vấn đề, thay \(a=b=c\) hai vế cho kết quả khác nhau
Ta sẽ chứng minh BĐT mạnh hơn sau:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3+3\sqrt{\dfrac{3\left(a+b+c\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2\left(ab+bc+ca\right)^2}}=3+3\sqrt{Q}\)
Do \(\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\)
\(\Rightarrow Q\le\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{\left(a+b+c\right)\left(ab+bc+ca\right)}{abc}-3\ge3\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}}\)
\(\Leftrightarrow\dfrac{a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)}{abc}\ge3\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}}\)
\(\Leftrightarrow a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)\ge\dfrac{3}{\sqrt{2}}\sqrt{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(VT\ge3\sqrt[3]{abc\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)}\)
Do đó ta chỉ cần chứng minh:
\(8\left(abc\right)^2\left[\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)^2\right]^2\ge\left(abc\right)^3\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^3\)
\(\Leftrightarrow8\left[\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\right]^2\ge abc\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^3\)
\(\Leftrightarrow\dfrac{1}{8}\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^4\ge abc\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^3\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) (hiển nhiên đúng theo AM-GM)