Bạn học delta chưa nhỉ, HSG chắc chắn là học rồi:vv

Ta cần chứng minh: \(\dfrac{a^2}{2}+b^2+c^2>ab+bc+ca\Leftrightarrow\dfrac{a^2}{2}+b^2+c^2-ab-bc-ca>0\Leftrightarrow\dfrac{a^2}{4}+b^2+c^2+ab+ca+2bc-3bc+\dfrac{a^2}{4}>0\) \(\Leftrightarrow\left(\dfrac{a}{2}+b+c\right)^2+\dfrac{a^2}{12}+\dfrac{a^2}{6}-3bc>0\Leftrightarrow\left(\dfrac{a}{2}+b+c\right)^2+\dfrac{a^2-36bc}{12}+\dfrac{a^2}{6}>0\) Mà \(a^3>36;abc=1\Rightarrow a^3>36abc\Rightarrow a^2>36bc\) 

\(\Rightarrow\left(\dfrac{a}{2}+b+c\right)^2+\dfrac{a^2-36bc}{12}+\dfrac{a^2}{6}>0\) luôn đúng

bạn có viết đề sai ko?

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(3=a+b+c\ge3\sqrt[3]{abc}\Rightarrow abc\le1\)

BĐT tương đương:

\(3\left(ab+bc+ca\right)\ge abc\left[\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+6\right]\)

\(\Leftrightarrow3\left(ab+bc+ca\right)\ge abc\left[15-2\left(ab+bc+ca\right)\right]\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(2abc+3\right)\ge15abc\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\left(2abc+3\right)^2\ge225\left(abc\right)^2\)

Do \(\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)=9abc\)

Nên ta chỉ cần chứng minh:

\(\left(2abc+3\right)^2\ge25abc\)

\(\Leftrightarrow\left(1-abc\right)\left(9-4abc\right)\ge0\) (luôn đúng với \(0< abc\le1\))

Dấu "=" xảy ra khi \(a=b=c=1\)