giúp mình câu :
cho ab+bc+ac = 1
GBT : a / √1 +a ² + b / √1+b ² + c/ 11+c ² ≤ 3/2
0 bình luận về “giúp mình câu :
cho ab+bc+ac = 1
GBT : a / √1 +a ² + b / √1+b ² + c/ 11+c ² ≤ 3/2”
$\begin{array}{l}\underline{\text{Đáp án+Giải thích các bước giải:}}\\Đặt \,\, A=\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\\+)a^2+1\\=a^2+ab+bc+ca\\=a(a+b)+c(a+b)\\=(a+b)(a+c)\\→\sqrt{a^2+1}=\sqrt{(a+b)(a+c)}\\CMTT:\sqrt{b^2+1}=\sqrt{(b+c)(b+a)}\\\sqrt{c^2+1}=\sqrt{(c+a)(c+b)}\\→A=\dfrac{a}{\sqrt{(a+b)(a+c)}}+\dfrac{b}{\sqrt{(b+c)(b+a)}}+\dfrac{c}{(c+a)(c+b)}\\\text{Áp dụng BĐT cosi ta có}\\\dfrac{\sqrt{a}.\sqrt{a}}{\sqrt{(a+b)(a+c)}} \leq \dfrac{1}{2}(\dfrac{a}{a+b}+\dfrac{a}{a+c})\\CMTT:\dfrac{b}{\sqrt{b^2+1}} \leq \dfrac{1}{2}(\dfrac{b}{a+b}+\dfrac{b}{b+c})\\\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{1}{2}(\dfrac{c}{c+a}+\dfrac{c}{c+b})\\→A \leq \dfrac{1}{2}(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a})\\→A \leq \dfrac{3}{2}(ĐPCM)\\\text{Dấu “=” xảy ra khi}\\a=b=c=\sqrt{\dfrac{1}{3}}\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$
$\begin{array}{l}\underline{\text{Đáp án+Giải thích các bước giải:}}\\Đặt \,\, A=\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\\+)a^2+1\\=a^2+ab+bc+ca\\=a(a+b)+c(a+b)\\=(a+b)(a+c)\\→\sqrt{a^2+1}=\sqrt{(a+b)(a+c)}\\CMTT:\sqrt{b^2+1}=\sqrt{(b+c)(b+a)}\\\sqrt{c^2+1}=\sqrt{(c+a)(c+b)}\\→A=\dfrac{a}{\sqrt{(a+b)(a+c)}}+\dfrac{b}{\sqrt{(b+c)(b+a)}}+\dfrac{c}{(c+a)(c+b)}\\\text{Áp dụng BĐT cosi ta có}\\\dfrac{\sqrt{a}.\sqrt{a}}{\sqrt{(a+b)(a+c)}} \leq \dfrac{1}{2}(\dfrac{a}{a+b}+\dfrac{a}{a+c})\\CMTT:\dfrac{b}{\sqrt{b^2+1}} \leq \dfrac{1}{2}(\dfrac{b}{a+b}+\dfrac{b}{b+c})\\\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{1}{2}(\dfrac{c}{c+a}+\dfrac{c}{c+b})\\→A \leq \dfrac{1}{2}(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a})\\→A \leq \dfrac{3}{2}(ĐPCM)\\\text{Dấu “=” xảy ra khi}\\a=b=c=\sqrt{\dfrac{1}{3}}\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$