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Áp dụng BĐT Cauchy-Schwarz ta có:
\(P=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
\(=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\frac{2a}{a+b}\cdot\frac{2a}{a+c}}+\sqrt{\frac{2b}{a+b}\cdot\frac{b}{2\left(b+c\right)}}+\sqrt{\frac{2c}{a+c}\cdot\frac{c}{2\left(b+c\right)}}\)
\(\le\frac{1}{2}\left(\frac{2a}{a+b}+\frac{2b}{a+b}+\frac{2a}{a+c}+\frac{2c}{a+c}+\frac{b}{2\left(b+c\right)}+\frac{c}{2\left(b+c\right)}\right)\)
\(=\frac{1}{2}\left(2+2+\frac{1}{2}\right)=\frac{9}{4}\)
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)
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
Ta có: \(a^2-ab+3b^2+1=\left(a^2-2ab+b^2\right)+ab+\left(b^2+1\right)+b^2\)
\(=\left(a-b\right)^2+ab+\left(b^2+1\right)+b^2\ge ab+2b+b^2\)
\(=b\left(a+b+2\right)\Rightarrow\frac{1}{\sqrt{a^2-ab+3b^2+1}}\le\frac{1}{\sqrt{b\left(a+b+2\right)}}\)(1)
Tương tự: \(\frac{1}{\sqrt{b^2-bc+3c^2+1}}\le\frac{1}{\sqrt{c\left(b+c+2\right)}}\)(2); \(\frac{1}{\sqrt{c^2-ca+3a^2+1}}\le\frac{1}{\sqrt{a\left(c+a+2\right)}}\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3) và sử dụng AM - GM kết hợp liên tục BĐT \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\), ta được:
\(P\le\frac{1}{\sqrt{b\left(a+b+2\right)}}+\frac{1}{\sqrt{c\left(b+c+2\right)}}+\frac{1}{\sqrt{a\left(c+a+2\right)}}\)
\(=\Sigma\frac{2}{\sqrt{4b\left(a+b+2\right)}}\)\(\le\Sigma\left(\frac{1}{4b}+\frac{1}{a+b+2}\right)\)(AM - GM)
\(=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\text{}\Sigma\left(\frac{1}{a+b+2}\right)\)
\(\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\text{}\Sigma\left[\frac{1}{4}\left(\frac{1}{a+b}\right)+\frac{1}{2}\right]\)
\(\le\frac{3}{4}+\text{}\left[\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\text{}\Sigma\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}\right)\right]\)
\(=\frac{3}{4}+\text{}\left[\frac{3}{8}+\text{}\frac{1}{8}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]\le\frac{3}{4}+\frac{3}{8}+\frac{3}{8}=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
Dòng thứ 10 sửa lại cho mình là \(\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\Sigma\left[\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{2}\right)\right]\)
Do olm có lỗi là mỗi lần bấm dấu ngoặc là số nó tự động nhảy ra ngoài
Tìm điều gì của M bạn?
Mình nghĩ là tìm Min, Max \(M=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\).
Tìm Min: Ta có \(M^2\ge a+b+b+c+c+a=2\left(a+b+c\right)\ge2\sqrt{a^2+b^2+c^2}=2\).
Do đó \(M\geq\sqrt{2}\).Đẳng thức xảy ra khi a = b = 0; c = 1.
Tìm Max: Ta có \(M\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\le\sqrt{6\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt{6\sqrt{3}}=\sqrt[4]{108}\).