Ta có \(a+b+c\le\sqrt{3}\)
\(\Rightarrow\left(a+b+c\right)^2\le3\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}\le1\)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ac\)
\(\Rightarrow1\ge ab+bc+ac\)
\(\Rightarrow\left\{\begin{matrix}1+a^2\ge a^2+ab+bc+ac\\1+b^2\ge b^2+ab+bc+ac\\1+c^2\ge c^2+ab+bc+ac\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\sqrt{1+a^2}\ge\sqrt{a^2+ab+bc+ca}\\\sqrt{1+b^2}\ge\sqrt{b^2+ab+bc+ca}\\\sqrt{1+c^2}\ge\sqrt{c^2+ab+bc+ca}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{1+a^2}}\le\frac{a}{\sqrt{a^2+ab+bc+ac}}\\\frac{b}{\sqrt{1+b^2}}\le\frac{b}{\sqrt{b^2+ab+bc+ac}}\\\frac{c}{\sqrt{1+c^2}}\le\frac{c}{\sqrt{c^2+ab+bc+ac}}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(\Rightarrow\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{a}{\sqrt{a\left(a+b\right)+c\left(a+b\right)}}+\frac{b}{\sqrt{b\left(b+a\right)+c\left(a+b\right)}}+\frac{c}{\sqrt{c\left(c+a\right)+b\left(c+a\right)}}\)
\(\Rightarrow\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Xét \(\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng bất đẳng thức Cauchy ngược dấu cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}\sqrt{\left(a+b\right)\left(a+c\right)}\ge\frac{2a+b+c}{2}\\\sqrt{\left(a+b\right)\left(b+c\right)}\ge\frac{a+2b+c}{2}\\\sqrt{\left(c+a\right)\left(c+b\right)}\ge\frac{a+b+2c}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{2a}{2b+b+c}\\\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\frac{2b}{a+2b+c}\\\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{2c}{a+b+2c}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le2\left(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\right)\)
Chứng minh rằng: \(2\left(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\right)\le\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\le\frac{3}{4}\)
Áp dụng bất đẳng thức \(\frac{1}{a+b}\ge\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\frac{a}{2a+b+c}=\frac{a}{a+c+a+b}\le\frac{a}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
\(\Rightarrow\frac{b}{a+2b+c}=\frac{b}{a+b+b+c}\le\frac{b}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\)
\(\Rightarrow\frac{c}{a+b+2c}=\frac{c}{a+c+b+c}\le\frac{c}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)
\(\Rightarrow VT\le\frac{a}{4\left(a+b\right)}+\frac{a}{4\left(a+c\right)}+\frac{b}{4\left(a+b\right)}+\frac{b}{4\left(b+c\right)}+\frac{c}{4\left(a+c\right)}+\frac{c}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\frac{a}{4\left(a+b\right)}+\frac{b}{4\left(a+b\right)}+\frac{a}{4\left(a+c\right)}+\frac{c}{4\left(a+c\right)}+\frac{b}{4\left(b+c\right)}+\frac{c}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\left(đpcm\right)\)
\(\Rightarrow2\left(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\right)\le\frac{3}{2}\)
\(\Rightarrow\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{3}{2}\)
Vậy \(\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{3}{2}\left(đpcm\right)\)
Lời giải khác:
Áp dụng bđt Cauchy-Schwarz:
\((a^2+1)(1+3)\geq (a+\sqrt{3})^2\)\(\Rightarrow \frac{a}{\sqrt{a^2+1}}\leq \frac{2a}{a+\sqrt{3}}\)
Thực hiện tương tự với các phân thức còn lại:
\(\Rightarrow \frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\leq 2\left ( \frac{a}{a+\sqrt{3}}+\frac{b}{b+\sqrt{3}}+\frac{c}{c+\sqrt{3}} \right )=2A\) $(1)$
Lại có:
\(\)\(A=\left ( 1-\frac{\sqrt{3}}{a+\sqrt{3}} \right )+\left ( 1-\frac{\sqrt{3}}{b+\sqrt{3}} \right )+\left ( 1-\frac{\sqrt{3}}{c+\sqrt{3}} \right )=3-\sqrt{3}\left ( \frac{1}{a+\sqrt{3}}+\frac{1}{b+\sqrt{3}}+\frac{1}{c+\sqrt{3}} \right )\)
Cauchy-Schwarz kết hợp với \(a+b+c\leq \sqrt{3}\):
\(A\leq 3-\frac{9\sqrt{3}}{a+b+c+3\sqrt{3}}\leq 3-\frac{9\sqrt{3}}{4\sqrt{3}}=\frac{3}{4}\) $(2)$
Từ \((1),(2)\Rightarrow \text{VT}\leq 2A\leq \frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Cách khác nữa:
Nhớ là \(f\left(x\right)=\frac{x}{\sqrt{x^2+1}}\) là 1 hàm lõm khi x>0, điều này xảy ra khi
\(f''(x)=-\dfrac{3x}{(x^2+1)^{\frac{5}{2}}}<0\). giờ thì sử dụng BĐT jensen
\(f\left(a\right)+f\left(b\right)+f\left(c\right)\le3f\left(\frac{a+b+c}{3}\right)=3f\left(\frac{\sqrt{3}}{3}\right)=\frac{3}{2}\left(a+b+c=\sqrt{3}\right)\)
Đạt dc GTLN khi \(a=b=c\).
Cách khác nữa:
Áp dụng BĐT AM-GM và BĐT W-P-M
\(\Sigma\frac{a}{\sqrt{a^2+1}}=\Sigma\frac{a}{\sqrt{a^2+3\cdot\frac{1}{3}}}\le\Sigma\frac{a}{\sqrt{4\sqrt[4]{\frac{a^2}{27}}}}=\frac{3\sqrt[8]{27}}{2}\cdot\frac{\Sigma\sqrt[4]{a^3}}{3}\)
\(\le\frac{3\sqrt[8]{27}}{2}\left(\frac{a+b+c}{3}\right)^{\frac{3}{4}}\le\frac{3\sqrt[8]{27}}{2}\cdot\left(\frac{\sqrt{3}}{3}\right)^{\frac{3}{4}}=\frac{3}{2}\) (xong)
P/s:Mong các bạn theo dõi và đóng góp ý kiến cho cách này
Ta có \(a+b+c\le\sqrt{3}\)
\(\Rightarrow\left(a+b+c\right)^2\le3\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}\le1\)
Theo hệ quả của bất đẳng thức Cauchy\
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ca\)
\(\Rightarrow1\ge ab+bc+ca\)
\(\Rightarrow\left\{\begin{matrix}1+a^2\ge a^2+ab+bc+ca=\left(a+c\right)\left(a+b\right)\\1+b^2\ge b^2+ab+bc+ca=\left(b+c\right)\left(a+b\right)\\1+c^2\ge c^2+ab+bc+ca=\left(b+c\right)\left(c+a\right)\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\sqrt{1+a^2}\ge\sqrt{\left(a+c\right)\left(a+b\right)}\\\sqrt{1+b^2}\ge\sqrt{\left(b+c\right)\left(a+b\right)}\\\sqrt{1+c^2}\ge\sqrt{\left(b+c\right)\left(c+a\right)}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{a^2+1}}\le\frac{a}{\sqrt{\left(a+c\right)\left(a+b\right)}}\\\frac{b}{\sqrt{1+b^2}}\le\frac{b}{\sqrt{\left(b+c\right)\left(a+b\right)}}\\\frac{c}{\sqrt{1+c^2}}\le\frac{c}{\sqrt{\left(b+c\right)\left(c+a\right)}}\end{matrix}\right.\)
\(\Rightarrow VT\le\frac{a}{\sqrt{\left(a+c\right)\left(a+b\right)}}+\frac{b}{\sqrt{\left(b+c\right)\left(a+b\right)}}+\frac{c}{\sqrt{\left(b+c\right)\left(c+a\right)}}\)
\(\Leftrightarrow VT\le\sqrt{\frac{a^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\frac{b^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\frac{c^2}{\left(b+c\right)\left(c+a\right)}}\) ( 1 )
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left\{\begin{matrix}2\sqrt{\frac{a^2}{\left(a+c\right)\left(a+b\right)}}\le\frac{a}{a+c}+\frac{a}{a+b}\\2\sqrt{\frac{b^2}{\left(b+c\right)\left(a+b\right)}}\le\frac{b}{b+c}+\frac{b}{a+b}\\2\sqrt{\frac{c^2}{\left(b+c\right)\left(c+a\right)}}\le\frac{c}{b+c}+\frac{c}{c+a}\end{matrix}\right.\)
\(\Rightarrow2\left(\sqrt{\frac{a^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\frac{b^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\frac{c^2}{\left(b+c\right)\left(c+a\right)}}\right)\le\frac{a}{a+b}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{c}{c+a}+\frac{a}{a+c}\)
\(\Rightarrow2\left(\sqrt{\frac{a^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\frac{b^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\frac{c^2}{\left(b+c\right)\left(c+a\right)}}\right)\le1+1+1=3\)
\(\Rightarrow\sqrt{\frac{a^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\frac{b^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\frac{c^2}{\left(b+c\right)\left(c+a\right)}}\le\frac{3}{2}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{3}{2}\) ( đpcm )
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