Áp dụng BĐT Cauchy-Schwarz ta có:

\(P=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)

\(=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(=\sqrt{\frac{2a}{a+b}\cdot\frac{2a}{a+c}}+\sqrt{\frac{2b}{a+b}\cdot\frac{b}{2\left(b+c\right)}}+\sqrt{\frac{2c}{a+c}\cdot\frac{c}{2\left(b+c\right)}}\)

\(\le\frac{1}{2}\left(\frac{2a}{a+b}+\frac{2b}{a+b}+\frac{2a}{a+c}+\frac{2c}{a+c}+\frac{b}{2\left(b+c\right)}+\frac{c}{2\left(b+c\right)}\right)\)

\(=\frac{1}{2}\left(2+2+\frac{1}{2}\right)=\frac{9}{4}\)

cảm ơn nha

Áp dụng BĐT Cauchy-Schwarz ta có :

\(P=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)

\(=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(=\sqrt{\frac{2a}{a+b}.\frac{2a}{a+c}}+\sqrt{\frac{2b}{a+b}.\frac{b}{2\left(b+c\right)}}+\sqrt{\frac{2c}{a+c}.\frac{c}{2\left(b +c\right)}}\)

\(\le\frac{1}{2}\left(\frac{2a}{a+b}+\frac{2b}{a+b}+\frac{2a}{a+c}+\frac{2c}{a+c}+\frac{b}{2\left(b+c\right)}+\frac{c}{2\left(b+c\right)}\right)\)

\(=\frac{1}{2}\left(2+2+\frac{1}{2}\right)=\frac{9}{4}\)

P/s : Mình tự nghĩ chứ không phải mình copy đâu

Cần c/m: \(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge3\sqrt{2}\)

Mặt khác \(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\left(\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}\right)\ge9\)

Nên ta chỉ cần c/m  \(P=\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}\le\frac{9}{3\sqrt{2}}=\frac{3\sqrt{2}}{2}\)

Ta có

\(P.\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{\left(a+b\right).2}}+\frac{1}{\sqrt{\left(b+c\right).2}}+\frac{1}{\sqrt{\left(c+a\right).2}}\)

\(=\sqrt{\frac{1}{a+b}}.\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{b+c}}.\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{c+a}}.\sqrt{\frac{1}{2}}\)

\(\le\frac{1}{2}\left(\frac{1}{a+b}+\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{b+c}+\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{c+a}+\frac{1}{2}\right)\)

\(=\frac{1}{2}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)+\frac{3}{4}\le\frac{1}{8}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)+\frac{3}{4}\)

\(=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{3}{4}=\frac{1}{4}.3+\frac{3}{4}=\frac{3}{2}\)

Suy ra  \(P\le\frac{3}{2}:\frac{1}{\sqrt{2}}=\frac{3\sqrt{2}}{2}\)

BĐT được c/m

Đẳng thức xảy ra  \(\Leftrightarrow a=b=c=1\)

\(a^2+1=ab+bc+ca+a^2=\left(a+b\right)\left(a+c\right)\)

tương tự \(\Rightarrow\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(\left(a+b\right)\left(c+a\right)\left(b+c\right)=a^2b+b^2a+c^2a+a^2c+b^2c+c^2b+2abc\)

\(\Rightarrow\)VT=\(a^2b+b^2a+b^2c+c^2b+c^2a+a^2c+3abc\) =\(ab\left(a+b\right)+bc\left(a+b\right)+ca\left(a+b\right)+c\left(ab+bc+ca\right)\)=a+b+c

ta có (a+b+c)^2>=3(ab+bc+ca)=3 nên a+b+c>=căn3(đccm)

Áp dụng bất đẳng thức Cosi, ta có:

\(\left(a^2+b+c\right)\left(1+b+c\right)\ge\left(a+b+c\right)^2\)Do đó, để chứng minh bất đẳng thức đã cho, ta chỉ cần chứng minh rằng:

\(\frac{a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}}{a+b+c}\le\sqrt{3}\)

Áp dụng bất đẳng thức Côsi lần thứ hai ta nhận được:

\(VT=\frac{\sqrt{a}\sqrt{a\left(1+b+c\right)}+\sqrt{b}\sqrt{b\left(1+c+a\right)}+\sqrt{c}\sqrt{c\left(1+a+b\right)}}{a+b+c}\)

\(\le\frac{\sqrt{\left(a+b+c\right)\left[a\left(1+b+c\right)+b\left(1+c+a\right)+c\left(1+a+b\right)\right]}}{a+b+c}\)

\(=\sqrt{1+\frac{2\left(ab+bc+ca\right)}{a+b+c}}\)

\(\le\sqrt{1+\frac{2\left(a+b+c\right)}{3}}\)

\(\le\sqrt{1+\frac{2\sqrt{3\left(a^2+b^2+c^2\right)}}{3}}=\sqrt{3}\left(đpcm\right)\)

Đẳng thức xảy ra khi và chỉ khi a = b = c = 1.

sửa đề thành \(a^2+b^2+c^2=3\) nhé

đặt \(A=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

\(=>A^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)

\(=>A^2\le\left[\left(\sqrt{a+b}\right)^2+\left(\sqrt{b+c}\right)^2+\left(\sqrt{c+a}\right)^2\right].3\)

\(=>A^2\le\left[2\left(a+b+c\right)\right]3=2.3=6\)

\(=>A\le\sqrt{6}\left(dpcm\right)\)

dấu"=" xảy ra<=>a=b=c=1/3

Ta có:\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2=\left(1.\sqrt{a+b}+1.\sqrt{b+c}+1.\sqrt{c+a}\right)^2\)

  \(\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)=3.2=6\)

\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)

Dấu "=" xảy ra <=> a=b=c=1/3

Áp dụng bđt \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\) 

\(\Rightarrow\sqrt{a^2+b^2}\ge\frac{a+b}{\sqrt{2}}\)

C/m tương tự \(\sqrt{b^2+c^2}\ge\frac{b+c}{\sqrt{2}}\)

                        \(\sqrt{a^2+c^2}\ge\frac{a+c}{\sqrt{2}}\)

Cộng 3 vế của 3 bđt trên lại được

\(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\frac{2\left(a+b+c\right)}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)

Dấu "=" tại a = b = c = ¹/₃

cảm ơn bạn nhiều nha