Question

Cho các số dương a; b; c thỏa mãn \(a+b+c\text{ ≤ }3\). CMR

\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{2009}{ab+bc+ca}\text{ ≥ }670\)

Answer 1

ta có:\(A=\dfrac{1}{a^2+b^2+c^2}+\dfrac{2009}{ab+bc+ca}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{4}{2\left(ab+bc+ca\right)}+\dfrac{2007}{ab+bc+ca}\)

Áp dụng BĐT cauchy-schwarz:

\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{4}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(1+2\right)^2}{\left(a+b+c\right)^2}\ge\dfrac{9}{9}=1\)

mà \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le3\)

do đó \(A\ge1+\dfrac{2007}{3}=670\)

dấu = xảy ra khi và chỉ khi a=b=c=1(làm tắt)

Answer 2

\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{2009}{ab+bc+ca}\)

\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{2007}{ab+bc+ca}\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{2007}{\left(a+b+c\right)^2}\ge670\)