Cho a, b, c là các số thực dương. Chứng minh rằng;
$\sqrt[3]{\frac{a^{2}+bc}{b^{2}+c^{2}}}+\sqrt[3]{\frac{b^{2}+ca}{c ^{2}+a^{2}}}+\sqrt[3]{\frac{c^{2}+ab}{a^{2}+b^{2}}}\geq \frac{9\sqrt[3]{abc}}{a+b+c}$.
Cho a, b, c là các số thực dương. Chứng minh rằng;
$\sqrt[3]{\frac{a^{2}+bc}{b^{2}+c^{2}}}+\sqrt[3]{\frac{b^{2}+ca}{c ^{2}+a^{2}}}+\sqrt[3]{\frac{c^{2}+ab}{a^{2}+b^{2}}}\geq \frac{9\sqrt[3]{abc}}{a+b+c}$.
"IF YOU HAVE A DREAM TO CHASE,NOTHING NOTHING CAN STOP YOU"_M10
Áp dụng bất đẳng thức Cô-si cho 3 số dương:
$\frac{a(b^{2}+c^{2})+b(a^{2}+c^{2})+c(a^{2}+b^{2})}{a^{2}+bc}=\frac{a(b^{2}+c^{2})}{a^{2}+bc}+b+c\geq 3\sqrt[3]{\frac{abc(b^{2}+c^{2})}{a^{2}+bc}}$
$=>\sqrt[3]{\frac{a^{2}+bc}{b^{2}+c^{2}}}\geq \frac{3(a^{2}+bc)\sqrt[3]{abc}}{a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2})}$
$=>VT=\sum \sqrt[3]{\frac{a^{2}+bc}{b^{2}+c^{2}}}\geq \frac{3(a^{2}+b^{2}+c^{2}+ab+bc+ac).\sqrt[3]{abc}}{a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2})}(*)$
KMTTQ, cho $a\geq b\geq c$
$(a+b+c)(a^{2}+b^{2}+c^{2}+ab+bc+ac)-3\left [ a(b^{2}+c^{2})+b(a^{2}+c^{2})+c(a^{2}+b^{2}) \right ]=a(a-b)^{2}+c(b-c)^{2}+(a-b+c)(a-b)(b-c)\geq 0$
$=>\frac{3(a^{2}+b^{2}+c^{2}+ab+bc+ac)}{a(b^{2}+c^{2})+b(a^{2}+c^{2})+c(a^{2}+b^{2})}\geq \frac{9}{a+b+c}(**)$
From $(*),(**)=>VT\geq VP (Q.E.D)$
Đẳng thức xảy ra khi $a=b=c$.
Treasure every moment that you have!
And remember that Time waits for no one.
Yesterday is history. Tomorrow is a mystery.
Today is a gift. That’s why it’s called the present.
0 thành viên, 0 khách, 0 thành viên ẩn danh