Lời giải:
\(\sum \frac{a^2+b^2}{2c}=\left (\frac{a^2}{2c}+\frac{b^2}{2a}+\frac{c^2}{2b}\right)+\left (\frac{a^2}{2b}+\frac{b^2}{2c}+\frac{c^2}{2a}\right)\)
Áp dụng BĐT AM-GM:
\(\left\{\begin{matrix} \frac{a^2}{2c}+\frac{c}{2}\geq 2\sqrt{\frac{a^2}{4}}=a\\ \frac{b^2}{2a}+\frac{a}{2}\geq 2\sqrt{\frac{b^2}{4}}=b\\ \frac{c^2}{2b}+\frac{b}{2}\geq 2\sqrt{\frac{c^2}{4}}=c\end{matrix}\right.\). Cộng theo vế và rút gọn:
\(\Rightarrow \frac{a^2}{2c}+\frac{b^2}{2a}+\frac{c^2}{2b}\geq \frac{a+b+c}{2}\)
Tương tự: \(\frac{a^2}{2b}+\frac{b^2}{2c}+\frac{c^2}{2a}\geq \frac{a+b+c}{2}\)
Do đó, \(\sum \frac{a^2+b^2}{2c}\geq a+b+c\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c\)
Cái bài này cũng bình thường :v
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT=\dfrac{a^2+b^2}{2c}+\dfrac{b^2+c^2}{2a}+\dfrac{c^2+a^2}{2b}\)
\(\ge\dfrac{\dfrac{\left(a+b\right)^2}{2}}{2c}+\dfrac{\dfrac{\left(b+c\right)^2}{2}}{2a}+\dfrac{\dfrac{\left(c+a\right)^2}{2}}{2b}\)
\(=\dfrac{\left(a+b\right)^2}{4c}+\dfrac{\left(b+c\right)^2}{4a}+\dfrac{\left(c+a\right)^2}{4b}\)
\(\ge\dfrac{\left(2a+2b+2c\right)^2}{4a+4b+4c}=\dfrac{4\left(a+b+c\right)^2}{4\left(a+b+c\right)}=a+b+c=VP\)
Khi \(a=b=c\)
Ta có:
\(\dfrac{a^2}{2c}+\dfrac{b^2}{2c}+\dfrac{b^2}{2a}+\dfrac{c^2}{2a}+\dfrac{c^2}{2b}+\dfrac{a^2}{2b}\)
\(\ge\dfrac{\left(2a+2b+2c\right)^2}{4a+4b+4c}=a+b+c\)
Dấu = xảy ra khi a = b = c
