Cho ba số dương a,b,c thỏa mãn \(a^2+b^2+c^2=1\) .Chứng minh rằng:
\(\frac{a^2}{1+b-a}+\frac{b^2}{1+c-b}+\frac{c^2}{1+a-c}\ge1\)
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\(sigma\frac{a}{1+b-a}=sigma\frac{a^2}{a+ab-a^2}\ge\frac{\left(a+b+c\right)^2}{a+b+c+\frac{\left(a+b+c\right)^2}{3}-\frac{\left(a+b+c\right)^2}{3}}=1\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
\(\frac{1}{b^2+c^2}=\frac{1}{1-a^2}=1+\frac{a^2}{b^2+c^2}\le1+\frac{a^2}{2bc}\)
Tương tự cộng lại quy đồng ta có đpcm
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
ap dung bdt am gm
\(\sqrt{1+8a^3}=\sqrt{\left(1+2a\right)\left(4a^2-4a+1\right)}\)\(\le\frac{1+2a+4a^2-2a+1}{2}=\frac{4a^2+2}{2}=2a^2+1\)
\(\Rightarrow\frac{1}{\sqrt{1+8a^3}}\ge\frac{1}{2a^2+1}\)
tuongtu ta cung co \(\frac{1}{\sqrt{1+8b^3}}\ge\frac{1}{2b^2+1};\frac{1}{\sqrt{1+8c^3}}\ge\frac{1}{2c^2+1}\)
\(\Rightarrow\)VT\(\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\)
tiep tuc ap dung bat cauchy-schwarz dang engel ta co
\(VT\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\ge\frac{\left(1+1+1\right)^2}{2\left(a^2+b^2+c^2\right)+3}=\frac{3^2}{6+3}=1\)(dpcm)
dau = xay ra \(\Leftrightarrow a=b=c=1\)
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\(ab+bc+ac=3abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
Áp dụng BĐT Cauchy, ta có : \(a^2+1\ge2a\Rightarrow\frac{1}{a^2+1}\le\frac{1}{2a}\)
Tương tự : \(\frac{1}{b^2+1}\le\frac{1}{2b}\) ; \(\frac{1}{c^2+1}\le\frac{1}{2c}\)
Cộng theo vế được :
\(P=\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\le\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
Vậy maxP = 3/2 tại a = b = c = 1
\(1-\frac{a^2b}{2+a^2b}\ge1-\frac{a^2b}{3.\sqrt[3]{a^2b}}\)\(\rightarrow1-3\sqrt[3]{a^4b^2}=3.\sqrt[3]{ab.ab.a^2}\rightarrow.....\)
BĐT cần chứng minh tương đương với \(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le1\)
Áp dụng BĐT Cauchy ta có: \(2+a^2b=1+1+a^2b\ge3\sqrt[3]{a^2b}\)
Do đó ta được \(\frac{a^2b}{1+a^2b}\le\frac{a^2b}{3\sqrt[3]{a^2b}}=\frac{a\sqrt[3]{ab^2}}{3}\)
Hoàn toàn tương tự ta được \(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le\frac{a\sqrt[3]{ab^2}+b\sqrt[3]{bc^2}+c\sqrt[3]{ca}}{3}\)
Cũng theo BĐT Cauchy ta được \(\sqrt[3]{ab^2}\le\frac{a+b+b}{3}=\frac{a+2b}{3}\)
\(\Rightarrow a\sqrt[3]{ab^2}\le\frac{a\left(a+2b\right)}{3}=\frac{a^2+2ab}{3}\)
Tương tự cũng được \(a\sqrt[3]{ab^2}+b\sqrt[3]{bc^2}+c\sqrt[3]{ca}\le\frac{\left(a+b+c\right)^2}{3}=3\)
Từ đó ta được\(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le1\)
Vậy BĐT được chứng minh. Dấu "=" xảy ra <=> a=b=c=1
1,
\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)
Áp dụng bất đẳng thức Cauchy-Schwarz dạng phân thức, ta được: \(VT=\frac{a^4}{a^2+a^2b-a^3}+\frac{b^4}{b^2+b^2c-b^3}+\frac{c^4}{c^2+c^2a-c^3}\)\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)+\left(a^2b+b^2c+c^2a\right)-\left(a^3+b^3+c^3\right)}\) \(=\frac{1}{1+\left(a^2b+b^2c+c^2a\right)-\left(a^3+b^3+c^3\right)}\)
Ta cần chứng minh \(\frac{1}{1+\left(a^2b+b^2c+c^2a\right)-\left(a^3+b^3+c^3\right)}\ge1\)hay \(a^3+b^3+c^3\ge a^2b+b^2c+c^2a\)
Đây là bất đẳng thức quen thuộc có nhiều cách chứng minh:
** Cách 1: Áp dụng AM - GM, ta được: \(a^3+a^3+b^3\ge3a^2b\); \(b^3+b^3+c^3\ge3b^2c\); \(c^3+c^3+a^3\ge3c^2a\)
Cộng từng vế ba bất đẳng thức trên
** Cách 2: Giả sử \(a\le b\le c\)
Có: \(a^3+b^3+c^3=a^2b+b^2c+c^2a+\left(c^2-a^2\right)\left(b-a\right)+\left(c^2-b^2\right)\left(c-b\right)\ge a^2b+b^2c+c^2a\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\).
Or the following SOS:![](data:image/png;base64,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)
* Hoặc mạnh hơn với a,b,c thực thỏa mãn \(a+b\ge0,b+c\ge0,c+a\ge0\)
\(a^3+b^3+c^3-a^2b-b^2c-c^2a\)
\(=\frac{\left(a^2+b^2-2c^2\right)^2+3\left(a^2-b^2\right)^2+\Sigma_{cyc}4\left(a+b\right)\left(c+a\right)\left(a-b\right)^2}{8\left(a+b+c\right)}\ge0\)