Cho a,b,c > 0 và a+b+c=3 CMR: $\frac{a}{b^2+1}$ + $\frac{b}{c^2+1}$ + $\frac{c}{a^2+1}$ $\geq$ 3 31/08/2021 Bởi Caroline Cho a,b,c > 0 và a+b+c=3 CMR: $\frac{a}{b^2+1}$ + $\frac{b}{c^2+1}$ + $\frac{c}{a^2+1}$ $\geq$ 3
Đáp án: $\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge \dfrac32$ Giải thích các bước giải: Ta có: $P=\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}$ $\to P=\left(a-\dfrac{ab^2}{b^2+1}\right)+\left(b-\dfrac{bc^2}{c^2+1}\right)+\left(c-\dfrac{ca^2}{a^2+1}\right)$ $\to P=\left(a+b+c\right)-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)$ $\to P\ge \left(a+b+c\right)-\left(\dfrac{ab^2}{2b}+\dfrac{bc^2}{2c}+\dfrac{ca^2}{2a}\right)$ $\to P\ge \left(a+b+c\right)-\dfrac12\left(ab+bc+ca\right)$ $\to P\ge \left(a+b+c\right)-\dfrac12\cdot \dfrac13\left(a+b+c\right)^2$ $\to P\ge 3-\dfrac12\cdot\dfrac13\cdot 3^2$ $\to P\ge \dfrac32$ Dấu = xảy ra khi $a=b=c=1$ Bình luận
Đáp án: $\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge \dfrac32$
Giải thích các bước giải:
Ta có:
$P=\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}$
$\to P=\left(a-\dfrac{ab^2}{b^2+1}\right)+\left(b-\dfrac{bc^2}{c^2+1}\right)+\left(c-\dfrac{ca^2}{a^2+1}\right)$
$\to P=\left(a+b+c\right)-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)$
$\to P\ge \left(a+b+c\right)-\left(\dfrac{ab^2}{2b}+\dfrac{bc^2}{2c}+\dfrac{ca^2}{2a}\right)$
$\to P\ge \left(a+b+c\right)-\dfrac12\left(ab+bc+ca\right)$
$\to P\ge \left(a+b+c\right)-\dfrac12\cdot \dfrac13\left(a+b+c\right)^2$
$\to P\ge 3-\dfrac12\cdot\dfrac13\cdot 3^2$
$\to P\ge \dfrac32$
Dấu = xảy ra khi $a=b=c=1$