CMR với a,b>=1 thì $\frac{1}{a^2+1}$ +$\frac{1}{b^2+1}$ $\geq$ $\frac{2}{ab+1}$ 27/08/2021 Bởi aihong CMR với a,b>=1 thì $\frac{1}{a^2+1}$ +$\frac{1}{b^2+1}$ $\geq$ $\frac{2}{ab+1}$
Ta có: Xét hiệu $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1} – \dfrac{2}{ab+1}$ $=\dfrac{a^2+b^2+2}{(a^2+1)(b^2+1)}-\dfrac{2}{ab+1}$ $=\dfrac{(a^2+b^2+2)(ab+1)-2(a^2+1)(b^2+1)}{(a^2+1)(b^2+1)(ab+1)}$ $=\dfrac{a^3b+b^3a+2ab+a^2+b^2+2-2(ab)^2-a^2-b^2-2}{(a^2+1)(b^2+1)(ab+1)}$ $=\dfrac{ab(a^2+b^2)+2ab-2(ab)^2-2}{(a^2+1)(b^2+1)(ab+1)}$ $=\dfrac{ab(a^2-2ab+b^2)+2(ab-1)}{(a^2+1)(b^2+1)(ab+1)}$ $=\dfrac{ab(a-b)^2+2(ab-1)}{(a^2+1)(b^2+1)(ab+1)}$ Ta thấy: $a^2+1;b^2+1>0∀a;b$; $a;b≥1⇒ab≥1⇒ab+1≥2>0$ $⇒(a^2+1)(b^2+1)(ab+1)>0(1)$ Mặt khác: $(a-b)^2≥0⇒ab(a-b)^2≥0;ab≥1$ và $ab≥1⇒ab-1≥0⇒2(ab-1)≥0$ $⇒ab(a-b)^2+2(ab-1)≥0∀a;b(2)$ Từ $1;2⇒\dfrac{ab(a-b)^2+2(ab-1)}{(a^2+1)(b^2+1)(ab+1)}≥0$ $⇒\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1} – \dfrac{2}{ab+1}≥0$ Hay $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}≥\dfrac{2}{ab+1}$ Dấu $=$ xảy ra $⇔a-b=0;ab=1⇔a=b=1$ Bình luận
Đáp án: Ta có : $\frac{1}{a^{2}+1}$ + $\frac{1}{b^{2}+1}$ ≥ $\frac{2}{ab+1}$ ⇔ $\frac{a^{2}+1 + b^{2} + 1}{(a^{2}+1)(b^{2} + 1)}$ ≥ $\frac{2}{ab+1}$ ⇔($a^{2}$+ $b^{2}$+ 2)(ab+1) ≥ 2.(2.$a^{2}$.$b^{2}$+ $a^{2}$ + $b^{2}$+ 1) ⇔ (ab+1)($a^{2}$+ $b^{2}$) + 2ab ≥ 2( 2.$a^{2}$.$b^{2}$+ $a^{2}$+ $b^{2}$) ⇔ (ab+1)($a^{2}$+ $b^{2}$) + 2ab – 2(2.$a^{2}$.$b^{2}$+$a^{2}$+ $b^{2}$) ≥ 0 ⇔ ab($a^{2}$- 2ab+ $b^{2}$) – ( $a^{2}$- 2ab + $b^{2}$) ≥ 0 ⇔ ab.$(a-b)^{2}$ – .$(a-b)^{2}$ ≥ 0 ⇔ $(a-b)^{2}$(ab-1) ≥ 0 ( với mọi a,b ≥ 1) => đpcm $\huge\text{xin hay nhất ạ (^)__(^)}$ Bình luận
Ta có:
Xét hiệu $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1} – \dfrac{2}{ab+1}$
$=\dfrac{a^2+b^2+2}{(a^2+1)(b^2+1)}-\dfrac{2}{ab+1}$
$=\dfrac{(a^2+b^2+2)(ab+1)-2(a^2+1)(b^2+1)}{(a^2+1)(b^2+1)(ab+1)}$
$=\dfrac{a^3b+b^3a+2ab+a^2+b^2+2-2(ab)^2-a^2-b^2-2}{(a^2+1)(b^2+1)(ab+1)}$
$=\dfrac{ab(a^2+b^2)+2ab-2(ab)^2-2}{(a^2+1)(b^2+1)(ab+1)}$
$=\dfrac{ab(a^2-2ab+b^2)+2(ab-1)}{(a^2+1)(b^2+1)(ab+1)}$
$=\dfrac{ab(a-b)^2+2(ab-1)}{(a^2+1)(b^2+1)(ab+1)}$
Ta thấy: $a^2+1;b^2+1>0∀a;b$; $a;b≥1⇒ab≥1⇒ab+1≥2>0$
$⇒(a^2+1)(b^2+1)(ab+1)>0(1)$
Mặt khác: $(a-b)^2≥0⇒ab(a-b)^2≥0;ab≥1$ và $ab≥1⇒ab-1≥0⇒2(ab-1)≥0$
$⇒ab(a-b)^2+2(ab-1)≥0∀a;b(2)$
Từ $1;2⇒\dfrac{ab(a-b)^2+2(ab-1)}{(a^2+1)(b^2+1)(ab+1)}≥0$
$⇒\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1} – \dfrac{2}{ab+1}≥0$
Hay $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}≥\dfrac{2}{ab+1}$
Dấu $=$ xảy ra $⇔a-b=0;ab=1⇔a=b=1$
Đáp án:
Ta có :
$\frac{1}{a^{2}+1}$ + $\frac{1}{b^{2}+1}$ ≥ $\frac{2}{ab+1}$
⇔ $\frac{a^{2}+1 + b^{2} + 1}{(a^{2}+1)(b^{2} + 1)}$ ≥ $\frac{2}{ab+1}$
⇔($a^{2}$+ $b^{2}$+ 2)(ab+1) ≥ 2.(2.$a^{2}$.$b^{2}$+ $a^{2}$ + $b^{2}$+ 1)
⇔ (ab+1)($a^{2}$+ $b^{2}$) + 2ab ≥ 2( 2.$a^{2}$.$b^{2}$+ $a^{2}$+ $b^{2}$)
⇔ (ab+1)($a^{2}$+ $b^{2}$) + 2ab – 2(2.$a^{2}$.$b^{2}$+$a^{2}$+ $b^{2}$) ≥ 0
⇔ ab($a^{2}$- 2ab+ $b^{2}$) – ( $a^{2}$- 2ab + $b^{2}$) ≥ 0
⇔ ab.$(a-b)^{2}$ – .$(a-b)^{2}$ ≥ 0
⇔ $(a-b)^{2}$(ab-1) ≥ 0 ( với mọi a,b ≥ 1)
=> đpcm
$\huge\text{xin hay nhất ạ (^)__(^)}$