Chứng minh a/b+b/c+c/a>=2(27q^2-9q+1)/(9q^2-2q+(1-3q)căn(q(1-3q)))+1/q-6

Bích Ngọc Huỳnh & erone - anotherway

Ta sẽ tìm hàm số \(f\left(q\right)\) sao cho
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge f(q) \forall a,b,c>0.\)

\(\Leftrightarrow \sum \dfrac{a}{b}+\sum \dfrac{b}{a}\ge 2f(q)+\sum \dfrac{b}{a}-\sum \dfrac{a}{b} \)

Or \(\sum ab(a+b)-2abc\cdot f(q)\ge (a-b)(b-c)(c-a)\)

Need to pro \(\sum ab(a+b)-2abc\cdot f(q)\ge \sqrt{(a-b)^2(b-c)^2(c-a)^2}.\)

Đặt \(p=a+b+c,q=ab+bc+ca,r=abc\)

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

\(p=1 \) have; \((q-3r)-2f(q)\cdot r \ge \sqrt{q^2+18qr-27r^2-4q^3-4r}\)

\(\Leftrightarrow\)\((27+k^2)r^2+2(2-kq-9q)r+4q^3 \ge 0\)

\(\Delta_r ‘=(2-kq-9q)^2-4q^3(27+k^2) \)

\(=q^2(1-4q)k^2+2q(9q-2)k+(9q-2)^2-108q^3\)

Cho\(\Delta_r ‘=0 \) tìm dc \(k=\dfrac{2-9q\pm 4\sqrt{q(1-3q)^3}}{q(1-4q)}.\)

Ta chọn \(k=\dfrac{2-9q+ 4\sqrt{q(1-3q)^3}}{q(1-4q)}\). do đó \(f(q)=\dfrac{k-3}{2}=\dfrac{1-6q+6q^2+ 2\sqrt{q(1-3q)^3}}{q(1-4q)}\)

Suy ra

\( 1-6q+6q^2+ 2\sqrt{q(1-3q)^3}=\left[2\sqrt{q(1-3q)^3}-2(9q^2-2q)\right]+(24q^2-10q+1)\\ \)

\(=2\cdot \dfrac{q(1-3q)^3-(9q^2-2q)^2}{\sqrt{q(1-3q)^3}+2(9q^2-2q)}+(4q-1)(6q-1)\\ \)

\(=2\cdot \dfrac{q(1-4q)(27q^2-9q+1)}{\sqrt{q(1-3q)^3}+2(9q^2-2q)}+(4q-1)(6q-1)\)

Vậy \(f(q)=\dfrac{2(27q^2-9q+1)}{\sqrt{q(1-3q)^3}+2(9q^2-2q)}+\dfrac{1-6q}{q}\)