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\(B=\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
\(B=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(b+c\right)\left(c+a\right)}}\)
\(B\le\frac{\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{c}{c+a}}{2}=\frac{3}{2}\Rightarrow B_{max}=\frac{3}{2}\)
\(\text{Dấu "=" xảy ra khi và chỉ khi:}a=b=c=\frac{1}{\sqrt{3}}\)
Ta có: \(5a^2+2ab+2b^2=4a^2+2ab+b^2+\left(a^2+b^2\right)\ge4a^2+2ab+b^2+2ab=\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Lại có: \(\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)
Tương tự cộng lại ta có: \(VT\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Theo BĐT Bunhiacopxki ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{3}\)
\(\Rightarrow VT\le\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\)
Dấu = xảy ra khi \(a=b=c=\sqrt{3}\)
Ta có:\(7\left(\frac{1}{a^2}+...\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+2015\)
Mà \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le2015\)=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{6045}\)
\(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+...\)
Mà \(\left(2+1\right)\left(2a^2+b^2\right)\ge\left(2a+b\right)^2\)(bất dẳng thức buniacoxki)
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
Lại có \(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
=> \(P\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\le\frac{\sqrt{6045}}{3}\)
Vậy \(MaxP=\frac{\sqrt{6045}}{3}\)khi \(a=b=c=\frac{\sqrt{6045}}{2015}\)
Ta có: \(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{bc}{a+b}+\frac{bc}{a+c}}{2}\)
\(\frac{ca}{\sqrt{b+ac}}=\frac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\frac{\frac{ca}{a+b}+\frac{ca}{b+c}}{2}\)
\(\frac{ab}{\sqrt{c+ab}}=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{ab}{a+c}+\frac{ab}{b+c}}{2}\)
Cộng 3 vế ta được: \(P\le\frac{\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{a+b}+\frac{ca}{b+c}+\frac{ab}{a+c}+\frac{ab}{b+c}}{2}\)
\(=\frac{\frac{c\left(a+b\right)}{a+b}+\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}}{2}=\frac{a+b+c}{2}=\frac{1}{2}\)
Vậy MinP = 1/2
\(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{a.1+bc}}=\frac{bc}{\sqrt{a\left(a+b+c\right)+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{bc}{a+b}+\frac{bc}{a+c}\)
Bài toán quy về 2 bài toán nhỏ hơn!
Cho các số dương ab + bc +ca = 1.
a) Tìm Max: \(M=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
(Lời giải tại: Câu hỏi của Nguyễn Linh Chi. Bài làm của anh Thắng, trong lời giải có phần giống với đề bên trên.)
b) Tìm Min: \(N=a^2+28b^2+28c^2\)
Có: \(N=\frac{1}{4}\left(2a-7b-7c\right)^2+\frac{63}{4}\left(b-c\right)^2+7\left(ab+bc+ca\right)\ge7\left(ab+bc+ca\right)=7\)
Từ đó tìm được \(P\le\frac{9}{4}-7=-\frac{19}{4}\)
Đẳng thức xảy ra khi \(a=\frac{7}{\sqrt{15}};b=c=\frac{1}{\sqrt{15}}\)
Với ab + bc + ca = 1 và áp dụng BĐT AM - GM, ta được:
\(\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)\(\frac{2a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(b+a\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
\(=\sqrt{\frac{2a}{a+b}.\frac{2a}{a+c}}+\sqrt{\frac{2b}{a+b}.\frac{b}{2\left(b+c\right)}}+\sqrt{\frac{2c}{a+c}.\frac{c}{2\left(b+c\right)}}\)
\(\le\frac{\frac{2a}{a+b}+\frac{2a}{a+c}}{2}+\frac{\frac{2b}{a+b}+\frac{b}{2\left(b+c\right)}}{2}+\frac{\frac{2c}{a+c}+\frac{c}{2\left(b+c\right)}}{2}\)
\(=\frac{\frac{2\left(a+b\right)}{a+b}+\frac{2\left(a+c\right)}{a+c}+\frac{b+c}{2\left(b+c\right)}}{2}=\frac{2+2+\frac{1}{2}}{2}=\frac{9}{4}\)(*)
Mặt khác, cũng theo AM - GM, ta có:
\(\frac{a^2}{2}+\frac{49b^2}{2}\ge2\sqrt{\frac{a^2}{2}.\frac{49b^2}{2}}=7ab\)(1)
\(\frac{a^2}{2}+\frac{49c^2}{2}\ge2\sqrt{\frac{a^2}{2}.\frac{49c^2}{2}}=7ac\)(2)
\(\frac{7}{2}\left(b^2+c^2\right)\ge\frac{7}{2}.2\sqrt{b^2c^2}=7bc\)(3)
Cộng theo từng vế của 3 BĐT (1), (2), (3), ta được:
\(\frac{2a^2+56b^2+56c^2}{2}\ge7\left(ab+bc+ca\right)=7\)
hay \(a^2+28b^2+28c^2\ge7\)(**)
Từ (*) và (**) suy ra \(P=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}-a^2-28b^2-28c^2\)
\(\le\frac{9}{4}-7=\frac{-19}{4}\)
Đẳng thức xảy ra khi \(a=\frac{7}{\sqrt{15}};b=c=\frac{1}{\sqrt{15}}\)
Có bất đẳng thức xy+zt≥x+zy+txy+zt≥x+zy+t với x,z≥0x,z≥0 ,y,t>0y,t>0
Giả sử cc lớn nhất trong các số a,b,ca,b,c thì c≥13c≥13
Do a,b,c≥0a,b,c≥0 nên
Ta có P2≥aa+1+bb+1+cc+1≥a+ba+b+2+cc+1P2≥aa+1+bb+1+cc+1≥a+ba+b+2+cc+1
Mà a+ba+b+2+cc+1−12=1−c3−c+c−12(c+1)=(1−c)(3c−1)(3−c)(2c+2)≥0
Ta có:
\(P=\frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ca+a^2}}\)
\(=\frac{1}{\sqrt{\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}}+\frac{1}{\sqrt{\frac{1}{4}\left(b+c\right)^2+\frac{3}{4}\left(b-c\right)^2}}+\frac{1}{\sqrt{\frac{1}{4}\left(c+a\right)^2+\frac{3}{4}\left(c-a\right)^2}}\)
\(\le2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(\le2.\frac{1}{4}.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\right)=20\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2019\)
\(\Leftrightarrow7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=20\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2019\)
\(\Rightarrow7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le\frac{20}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2019\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le6057\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\sqrt{673}\)
Ta có:
\(\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}\ge\sqrt{\left(2a+b\right)^2}=2a+b\)
\(\Rightarrow\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)
Tương tự: \(\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\frac{1}{9}\left(\frac{2}{b}+\frac{1}{c}\right)\) ; \(\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\frac{1}{9}\left(\frac{2}{c}+\frac{1}{a}\right)\)
Cộng vế với vế:
\(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\sqrt{673}\)
\(P_{max}=\sqrt{673}\) khi \(a=b=c=\frac{1}{\sqrt{673}}\)