Question

[Nguyen]

cho a,b,c>0 thỏa mãn ab+bc+ca=3.cmr:

\(P=\frac{a^2b+b^2c+c^2a}{3}+\sqrt[3]{9\left(a+b+c\right)}\ge4\)

Answer 1

[tthnew]

Hôm qua em không có online. Bài này căng não@@

Đặt \(p=a+b+c;q=ab+bc+ca;r=abc\Rightarrow q=3\) thì \(p^2\ge3q=9\Rightarrow p\ge3\)

Chú ý: \(-4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2=(a-b)^2 (b-c)^2 (c-a)^2 \geq 0\)

\(\Rightarrow\) \(1/27(-2p^3-2\sqrt{(p^2-3q)^3}+9pq) \leq r \leq 1/27(-2p^3+2\sqrt{(p^2-3q)^3}+9pq)\)

Hay là: \(\frac{1}{27}\left(-2p^3-2\sqrt{\left(p^2-9\right)^3}+27p\right)\le r\le\frac{1}{27}\left(-2p^3+2\sqrt{\left(p^2-9\right)^3}+27p\right)\)

Nếu \(a\ge b\ge c\Rightarrow a^2b+b^2c+c^2a\ge ab^2+bc^2+ca^2\)

\(\Rightarrow a^2b+b^2c+c^2a\ge\frac{1}{2}\Sigma ab\left(a+b\right)=\frac{1}{2}\left(pq-3r\right)=\frac{3}{2}\left(p-3r\right)\)

Do đó: \(P\ge\frac{1}{2}\left(p-3r\right)+\sqrt[3]{9p}\ge\frac{1}{2}\left(p-\frac{1}{27}\left(-2p^3+2\sqrt{\left(p^2-9\right)^3}+27p\right)\right)+3\)

\(\ge\frac{1}{27}p^3-\frac{1}{27}\sqrt{\left(p^2-9\right)^3}+3=f\left(p\right)\). Dễ thấy khi p tăng thì f(p) tăng.

Do đó f(p) đạt giá trị nhỏ nhất khi p đạt giá trị nhỏ nhất. Hay là: \(f\left(p\right)\ge f\left(3\right)=4=VP\)

Trường hợp còn lại tối về em đăng, đang bận!

Answer 2

[tthnew]

Nếu \(a\le b\le c\Rightarrow\left(a-b\right)\left(b-c\right)\left(a-c\right)\le0\)

\(\Rightarrow\left(a-b\right)\left(b-c\right)\left(a-c\right)=-\left|\left(a-b\right)\left(b-c\right)\left(a-c\right)\right|=-\sqrt{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)

\(=-\sqrt{-4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2}\)

---------------------------------------------------------------------------------------------------------

Chú ý: \(-4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2=(a-b)^2 (b-c)^2 (c-a)^2 \geq 0\)

\(\Rightarrow\) \(1/27(-2p^3-2\sqrt{(p^2-3q)^3}+9pq) \leq r \leq 1/27(-2p^3+2\sqrt{(p^2-3q)^3}+9pq)\)

Hay là: \(\frac{1}{27}\left(-2p^3-2\sqrt{\left(p^2-9\right)^3}+27p\right)\le r\le\frac{1}{27}\left(-2p^3+2\sqrt{\left(p^2-9\right)^3}+27p\right)\)

Ta có: \(2\left(a^2b+b^2c+c^2a\right)=\Sigma ab\left(a+b\right)+\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

\(=pq-3r-\sqrt{-4p^3r + p^2q^2 + 18pqr - 4q^3 - 27r^2}\)

\(=3p-3r-\sqrt{-4p^3r + 9p^2 + 54pr - 108 - 27r^2}\)

Do đó: \(a^2b+b^2c+c^2a\)​​\(=\frac{3p-3r-\sqrt{-4p^3r + 9p^2 + 54pr - 108 - 27r^2}}{2}\)

Do đó: \(P\)\(=\frac{3p-3r-\sqrt{-4p^3r + 9p^2 + 54pr - 108 - 27r^2}}{6}\)\(+\sqrt[3]{9p}\ge4\)

\(\Leftrightarrow\frac{3p-3r}{6}+\sqrt[3]{9p}\ge4+\)\(\frac{\sqrt{-4p^3r + 9p^2 + 54pr - 108 - 27r^2}}{6}\)

Or \(3p-3r+6\sqrt[3]{9p}-24\ge\)\(\sqrt{-4p^3r + 9p^2 + 54pr - 108 - 27r^2}\)

Vì: \(VT=3p-3r+6\sqrt[3]{9p}-24\ge3p-\frac{pq}{3}+18-24=0\)

Nên bất đẳng thức trên tương đương:

​​\(\left(3p-3r+6\sqrt[3]{9p}-24\right)^2\ge\) \(-4p^3r + 9p^2 + 54pr - 108 - 27r^2\)

Em chịu thua :( @Akai Haruma @Nguyễn Việt Lâm giúp em với ạ.