Cho a, b, c >0 t/m: \(a^2\left(\dfrac{a}{c}+1\right)+b^2\left(\dfrac{b}{c}+1\right)=3\)
Tìm \(Min\) \(H=\dfrac{a+c}{b^3+2}+\dfrac{b+c}{a^3+2}-2\sqrt{a+b+c}\)
Ace Legona, Phương An, Neet,.....
\(GT\Leftrightarrow\dfrac{a^2\left(a+c\right)+b^2\left(b+c\right)}{c}=3\)\(\Leftrightarrow3c=a^3+b^3+\left(a^2+b^2\right)c\)
\(VT=\dfrac{\left(a+c\right)\left(b^3+2\right)}{2\left(b^3+2\right)}-\dfrac{b^3\left(a+c\right)}{2\left(b^3+2\right)}+\dfrac{\left(b+c\right)\left(a^3+2\right)}{2\left(a^3+2\right)}-\dfrac{a^3\left(b+c\right)}{2\left(a^3+2\right)}-2\sqrt{a+b+c}\)
\(=\dfrac{a+b+2c}{2}-\dfrac{b^3\left(a+c\right)}{2\left(b^3+1+1\right)}-\dfrac{a^3\left(b+c\right)}{2\left(a^3+1+1\right)}-2\sqrt{a+b+c}\)
Áp dụng BĐT AM-GM:
\(b^3+1+1\ge3b\) ; \(a^3+1+1\ge3a\).
Do đó :\(VT\ge\dfrac{a+b+2c}{2}-\dfrac{b^2\left(a+c\right)+a^2\left(b+c\right)}{6}-2\sqrt{a+b+c}\)
Để ý rằng \(b^2\left(a+c\right)+a^2\left(b+c\right)=ab\left(a+b\right)+\left(a^2+b^2\right)c\le a^3+b^3+\left(a^2+b^2\right).c=3c\)
\(\Rightarrow VT\ge\dfrac{a+b+2c}{2}-\dfrac{3c}{6}-2\sqrt{a+b+c}=\dfrac{a+b+c}{2}-2\sqrt{a+b+c}\)
\(=\dfrac{\left(\sqrt{a+b+c}-2\right)^2-4}{2}\ge-2\).
Dấu = xảy ra khi a=b=1;c=2
lần +)F_'o nào mình muốn động não cx gặp bài bạn này pót lên '-' vi diệu v~ .-. nhưng thôi k chắc chắn lắm nên để mấy ae lm, chứ t lm mà sai là khổ thân bạn ý :v
