Em nghĩ đề là a chứ không phải 2a ;v
\(P=\dfrac{a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\\ =\dfrac{a}{\sqrt{ab+bc+ac+a^2}}+\dfrac{b}{\sqrt{ab+bc+ac+b^2}}+\dfrac{c}{\sqrt{ab+bc+ac+c^2}}\\ =\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\\ \le\left(\dfrac{a}{2\left(a+b\right)}+\dfrac{a}{2\left(a+c\right)}\right)+\left(\dfrac{b}{2\left(a+b\right)}+\dfrac{b}{2\left(b+c\right)}\right)+\left(\dfrac{c}{2\left(a+c\right)}+\dfrac{c}{2\left(b+c\right)}\right)\)
\(=\dfrac{2\left(a+b+c\right)}{8\left(a+b+c\right)}=\dfrac{1}{4}\)
Áp dụng bđt : \(\dfrac{1}{xy}\le\dfrac{\dfrac{1}{x^2}+\dfrac{1}{y^2}}{2}\)
Dấu "=" xảy ra khi a=b=c=1/căn 3
Dự đoán điểm rơi b=c=ka. Ta có:
\(P=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}}+\dfrac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng BĐT AM-GM: \(\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{a+b}+\dfrac{a}{a+c}\)
\(\dfrac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}}=\dfrac{b.\sqrt{\dfrac{2k}{k+1}}}{\sqrt{\left(b+c\right).\dfrac{2k\left(a+b\right)}{k+1}}}\le\dfrac{b}{2}\sqrt{\dfrac{2k}{k+1}}.\left(\dfrac{1}{b+c}+\dfrac{\left(k+1\right)}{2k\left(a+b\right)}\right)\)
\(\dfrac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\dfrac{c}{2}.\sqrt{\dfrac{2k}{k+1}}\left(\dfrac{1}{b+c}+\dfrac{k+1}{2k\left(a+c\right)}\right)\)
\(\Rightarrow VT\le\dfrac{a}{a+b}+\sqrt{\dfrac{k+1}{8k}}.\dfrac{b}{a+b}+\dfrac{a}{a+c}+\sqrt{\dfrac{k+1}{8k}}.\dfrac{c}{a+c}+\sqrt{\dfrac{k}{2k+2}}\)
Tìm k sao cho \(\sqrt{\dfrac{k+1}{8k}}=1\Rightarrow k=\dfrac{1}{7}\)
Do đó trình bày lại bài toán ngắn gọn như sau:
Áp dụng BĐT AM-GM:
\(VT=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{2b}{\sqrt{4\left(b+c\right).\left(b+a\right)}}+\dfrac{2c}{\sqrt{4\left(b+c\right).\left(a+b\right)}}\)
\(\le\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{4\left(b+c\right)}+\dfrac{b}{a+b}+\dfrac{c}{4\left(b+c\right)}+\dfrac{c}{a+c}\)
\(=1+1+\dfrac{1}{4}=\dfrac{9}{4}\)
Dấu = xảy ra khi \(a=7b=7c=\dfrac{7}{\sqrt{15}}\)
