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+ \(2a+b+c=\left(a+b\right)+\left(a+c\right)\)
\(\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\) ( theo AM-GM )
\(\Rightarrow\left(2a+b+c\right)^2\ge4\left(a+b\right)\left(a+c\right)\)
\(\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{4\left(a+b\right)\left(a+c\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow b=c\)
+ Tương tự : \(\frac{1}{\left(2b+c+a\right)^2}\le\frac{1}{4\left(a+b\right)\left(b+c\right)}\). Dấu "=" xảy ra <=> a = c
\(\frac{1}{\left(2c+a+b\right)^2}\le\frac{1}{4\left(a+c\right)\left(b+c\right)}\). Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Do đó : \(P\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(b+c\right)}+\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\)
\(\Rightarrow P\le\frac{1}{2}\cdot\frac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}\)\(=8abc\)
\(\Rightarrow P\le\frac{a+b+c}{16abc}\)
+ \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\). Dấu :=" xảy ra \(\Leftrightarrow a=b\)
\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\). Dấu "=" xảy ra <=> b = c
\(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\). Dấu "=" xảy ra <=> c = a
\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow3\ge\frac{a+b+c}{abc}\) \(\Rightarrow a+b+c\le3abc\)
\(\Rightarrow P\le\frac{3abc}{16abc}=\frac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Giải:
Ta có:
\(P=\dfrac{a^3}{\sqrt{1+b^2}}+\dfrac{b^3}{\sqrt{1+c^2}}+\dfrac{c^3}{\sqrt{1+a^2}}\)
\(\Leftrightarrow P+3=\dfrac{a^3}{\sqrt{1+b^2}}+b^2+\dfrac{b^3}{\sqrt{1+c^2}}+c^2\dfrac{c^3}{\sqrt{1+a^2}}+a^2\)
\(\Leftrightarrow P+\dfrac{6}{4\sqrt{2}}=\dfrac{a^3}{2\sqrt{1+b^2}}+\dfrac{a^2}{2\sqrt{1+b^2}}+\dfrac{1+b^2}{4\sqrt{2}}+\dfrac{b^3}{2\sqrt{1+c^2}}+\dfrac{b^2}{2\sqrt{1+c^2}}+\dfrac{1+c^2}{4\sqrt{2}}+\dfrac{c^3}{2\sqrt{1+a^2}}+\dfrac{c^2}{2\sqrt{1+a^2}}+\dfrac{1+a^2}{4\sqrt{2}}\)
\(\ge3\sqrt[3]{\dfrac{a^6}{16\sqrt{2}}}+3\sqrt[3]{\dfrac{b^6}{16\sqrt{2}}}+3\sqrt[3]{\dfrac{c^6}{16\sqrt{2}}}\)
\(\Rightarrow P+\dfrac{3}{2\sqrt{2}}\ge\dfrac{3}{2\sqrt[3]{2\sqrt{2}}}\left(a^2+b^2+c^2\right)=\dfrac{9}{2\sqrt[6]{8}}\)
\(\Rightarrow P\ge\dfrac{9}{2\sqrt[6]{2^3}}-\dfrac{3}{2\sqrt{2}}=\dfrac{9}{2\sqrt{2}}-\dfrac{3}{2\sqrt{2}}=\dfrac{3}{\sqrt{2}}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng BĐt cauchy-schwarz:(dạng phân thức + đa thức )
\(P=\dfrac{a^3}{\sqrt{1+b^2}}+\dfrac{b^3}{\sqrt{1+c^2}}+\dfrac{c^3}{\sqrt{1+a^2}}=\dfrac{a^4}{a\sqrt{1+b^2}}+\dfrac{b^4}{b\sqrt{1+c^2}}+\dfrac{c^4}{c\sqrt{1+a^2}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\sqrt{1+b^2}+b\sqrt{1+c^2}+c\sqrt{1+a^2}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\sqrt{\left(a^2+b^2+c^2\right)\left(3+a^2+b^2+c^2\right)}}=\dfrac{9}{\sqrt{18}}=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}\)
dấu = xảy ra khi a=b=c=1
Ta có :\(\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}=\dfrac{1}{\sqrt{\left(4a^2+4ab+b^2\right)+\left(a^2-2ab+b^2\right)}}\)
\(=\dfrac{1}{\sqrt{\left(2a+b\right)^2+\left(a-b\right)^2}}\le\dfrac{1}{\sqrt{\left(2a+b\right)^2}}=\dfrac{1}{2a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\) (Cosi)
Tương tự cộng lại ta được :
\(P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{3}\sqrt{3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)}=\dfrac{1}{\sqrt{3}}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\sqrt{3}\)
\(\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)\(\le\) \(\dfrac{1}{3}\sqrt{3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)}\) làm thế nào hả bn ?
Câu 3. Dự đoán dấu "=" khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Dùng phương pháp chọn điểm rơi thôi :)
LG
Áp dụng bđt Cô-si được \(a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow1\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{3}\ge\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{27}\ge a^2b^2c^2\)
\(\Rightarrow\frac{1}{\sqrt{27}}\ge abc\)
Khi đó :\(B=a+b+c+\frac{1}{abc}\)
\(=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\)
\(\ge4\sqrt[4]{abc.\frac{1}{9abc}}+\frac{8}{9.\frac{1}{\sqrt{27}}}\)
\(=4\sqrt[4]{\frac{1}{9}}+\frac{8\sqrt{27}}{9}=\frac{4}{\sqrt[4]{9}}+\frac{8}{\sqrt{3}}=\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}\)
Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Vậy .........
2, \(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\left[\frac{a^2}{b+c}+\frac{\left(b+c\right)}{4}\right]+\left[\frac{b^2}{a+c}+\frac{\left(a+c\right)}{4}\right]+\left[\frac{c^2}{a+b}+\frac{\left(a+b\right)}{4}\right]-\frac{\left(a+b+c\right)}{2}\)
Áp dụng BĐT AM-GM ta có:
\(A\ge2.\sqrt{\frac{a^2}{4}}+2.\sqrt{\frac{b^2}{4}}+2.\sqrt{\frac{c^2}{4}}-\frac{\left(a+b+c\right)}{2}\)
\(A\ge a+b+c-\frac{6}{2}\)
\(A\ge6-3\)
\(A\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a^2}{b+c}=\frac{b+c}{4}\Leftrightarrow4a^2=\left(b+c\right)^2\Leftrightarrow2a=b+c\)(1)
\(\frac{b^2}{a+c}=\frac{a+c}{4}\Leftrightarrow4b^2=\left(a+c\right)^2\Leftrightarrow2b=a+c\)(2)
\(\frac{c^2}{a+b}=\frac{a+b}{4}\Leftrightarrow4c^2=\left(a+b\right)^2\Leftrightarrow2c=a+b\)(3)
Lấy \(\left(1\right)-\left(3\right)\)ta có:
\(2a-2c=c+b-a-b=c-a\)
\(\Rightarrow2a-2c-c+a=0\)
\(\Leftrightarrow3.\left(a-c\right)=0\)
\(\Leftrightarrow a-c=0\Leftrightarrow a=c\)
Chứng minh tương tự ta có: \(\hept{\begin{cases}b=c\\a=b\end{cases}}\)
\(\Rightarrow a=b=c=2\)
Vậy \(A_{min}=3\Leftrightarrow a=b=c=2\)
Áp dụng BĐT Cauchy dạng Engel , ta có :
\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\) ≥ \(\dfrac{\left(1+1+1\right)^2}{a+b+c+1+1+1}=\dfrac{9}{a+b+c+3}\text{ ≥}\dfrac{9}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\)
\("="\text{⇔}a=b=c=1\)
Lời giải:
Ta có:
\(P=\frac{a^2+1}{b^2+1}+\frac{b^2+1}{c^2+1}+\frac{c^2+1}{a^2+1}\)
\(=a^2+1-\frac{b^2(a^2+1)}{b^2+1}+b^2+1-\frac{c^2(b^2+1)}{c^2+1}+c^2+1-\frac{a^2(c^2+1)}{a^2+1}\)
\(=a^2+b^2+c^2+3-\left(\frac{b^2(a^2+1)}{b^2+1}+\frac{c^2(b^2+1)}{c^2+1}+\frac{a^2(c^2+1)}{a^2+1}\right)(*)\)
Vì \(a,b,c\geq 0; a+b+c=1\Rightarrow 0\leq a,b,c\leq 1\)
\(\Rightarrow 0\leq a^2,b^2,c^2\leq 1\)
Do đó:
\(\frac{b^2(a^2+1)}{b^2+1}+\frac{c^2(b^2+1)}{c^2+1}+\frac{a^2(c^2+1)}{a^2+1}\geq \frac{b^2(a^2+1)}{2}+\frac{c^2(b^2+1)}{2}+\frac{a^2(c^2+1)}{2}(**)\)
Từ \((*);(**)\Rightarrow P\leq a^2+b^2+c^2+3-\frac{a^2+b^2+c^2+(a^2b^2+b^2c^2+c^2a^2)}{2}=3+\frac{a^2+b^2+c^2-(a^2b^2+b^2c^2+c^2a^2)}{2}\)
Mà: \(a^2+b^2+c^2-(a^2b^2+b^2c^2+c^2a^2)=(a+b+c)^2-[2(ab+bc+ac)+(a^2b^2+b^2c^2+c^2a^2]\)
\(=1-[2(ab+bc+ac)+(a^2b^2+b^2c^2+c^2a^2]\leq 1\) do \(a,b,c\geq 0\)
Suy ra \(P\leq 3+\frac{a^2+b^2+c^2-(a^2b^2+b^2c^2+c^2a^2)}{2}\leq 3+\frac{1}{2}=\frac{7}{2}\)
Vậy \(P_{\max}=\frac{7}{2}\Leftrightarrow (a,b,c)=(1,0,0)\) và hoán vị.