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Ta có:
1+a2 = ab+bc+ca+a2 = a(a+b)+c(a+b)=(a+b)(a+c)
Tương tự: 1+b2 = (b+c)(b+a)
1+c2 = (c+a)(c+b)
\(\Rightarrow\) P = \(2a\sqrt{\dfrac{1}{\left(a+b\right)\left(a+c\right)}}+2b\sqrt{\dfrac{1}{\left(b+c\right)\left(b+a\right)}}+2c\sqrt{\dfrac{1}{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng BĐT Cô-si ta có:
P\(\le\)\(a\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+b\left(\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{b+a}\right)+c\left(\dfrac{1}{4\left(c+b\right)}+\dfrac{1}{c+a}\right)\)\(\le\)\(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{4\left(b+c\right)}+\dfrac{b}{b+a}+\dfrac{c}{4\left(c+b\right)}+\dfrac{c}{c+a}\)
= \(\dfrac{1}{4}+2=\dfrac{9}{4}\)
\(\Rightarrow\)Pmin = \(\dfrac{9}{4}\)
Dấu "=" xảy ra\(\Leftrightarrow\) b=c=\(\dfrac{a}{7}\)=\(\dfrac{\sqrt{15}}{15}\) \(\Rightarrow\) a = \(\dfrac{7\sqrt{15}}{15}\)
\(P=\dfrac{bc}{\dfrac{a^2bc}{c}+\dfrac{a^2bc}{b}}+\dfrac{ca}{\dfrac{b^2ac}{a}+\dfrac{b^2ac}{c}}+\dfrac{ab}{\dfrac{c^2ab}{b}+\dfrac{c^2ab}{a}}=\dfrac{\left(bc\right)^2}{a^2b^2c+a^2bc^2}+\dfrac{\left(ca\right)^2}{b^2a^2c+b^2ac^2}+\dfrac{\left(ab\right)^2}{c^2a^2b+c^2ab^2}=\dfrac{\left(bc\right)^2}{ab+ac}+\dfrac{\left(ca\right)^2}{ba+bc}+\dfrac{\left(ab\right)^2}{ca+cb}\ge\dfrac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}\ge\dfrac{3\sqrt[3]{\left(abc\right)^2}}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 1
a)Bunhia:
\(\left(1+2\right)\left(b^2+2a^2\right)\ge\left(1.b+\sqrt{2}.\sqrt{2}a\right)^2=\left(b+2a\right)^2\)
b)\(ab+bc+ca=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bđt câu a
=>VT\(\ge\)\(\dfrac{b+2a}{\sqrt{3}ab}+\dfrac{c+2b}{\sqrt{3}bc}+\dfrac{a+2c}{\sqrt{3}ca}\)
\(\Leftrightarrow VT\ge\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{2}{a}=3=VP\)
Tự tìm dấu "="
Nguyễn Việt LâmMashiro ShiinaBNguyễn Thanh HằngonkingCẩm MịcFa CTRẦN MINH HOÀNGhâu DehQuân Tạ MinhTrương Thị Hải Anh
1/ Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel :
\(A\ge\frac{\left(a+b+c\right)^2}{3\left(a+b+c\right)}=\frac{a+b+c}{3}=\frac{3}{3}=1\)
Dấu "=" xảy ra <=> a=b=c=1
Ta có:(Sử dụng bdt cô-si) \(\frac{bc}{a^2b+a^2c}+\frac{b+c}{4bc}\ge2\sqrt{\frac{bc}{a^2\left(b+c\right)}.\frac{b+c}{4bc}}=2.\frac{1}{2a}=\frac{1}{a}\)
=> \(\frac{bc}{a^2b+a^2c}\ge\frac{1}{a}-\frac{b+c}{4bc}\)
Chứng minh tương tự:\(\frac{ca}{b^2a+b^2c}\ge\frac{1}{b}-\frac{c+a}{4ca}\);\(\frac{ab}{c^2a+c^2b}\ge\frac{1}{c}-\frac{a+b}{4ab}\)
Từ đó \(P\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\left(\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}\right)\)
Mà\(\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)=> \(P\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\ge9\)(do a+b+c<=1)=> \(P\ge\frac{1}{2}.9=\frac{9}{2}\)
Dấu '=' xảy ra <=> \(\hept{\begin{cases}a+b+c=1\\\frac{bc}{a^2b+a^2c}=\frac{b+c}{4bc}\\a,b,c>0\end{cases}};...\)
<=> \(a=b=c=\frac{1}{3}\)
Vậy\(MinP=\frac{9}{2}\)khi a=b=c=1/3