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\(2\ge a^2+b^2\ge2ab\Rightarrow ab\le1\)
Ta có:
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}-\frac{2}{1+ab}=\frac{a^2+b^2+2}{a^2b^2+a^2+b^2+1}-\frac{2}{1+ab}\)
\(=\frac{\left(ab+1\right)\left(a^2+b^2+2\right)-2a^2b^2-a^2-b^2-2}{\left(1+ab\right)\left(1+a^2\right)\left(1+b^2\right)}=\frac{ab\left(a^2+b^2\right)-2a^2b^2+2ab-a^2-b^2}{\left(1+ab\right)\left(1+a^2\right)\left(1+b^2\right)}\)
\(=\frac{ab\left(a^2+b^2-2ab\right)-\left(a-b\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}=\frac{\left(ab-1\right)\left(a-b\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\le0;\forall ab\le1\)
\(\Rightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\le\frac{2}{1+ab}\)
Dấu "=" xảy ra khi \(a=b\)
\(1)\) \(\frac{1}{a^3+5}+\frac{1}{b^3+5}\le\frac{1}{3a+3}+\frac{1}{3b+3}=\frac{1}{3}\left(\frac{1}{a+1}+\frac{1}{b+1}\right)=\frac{1}{3}\left[\frac{a+b+2}{\left(a+1\right)\left(b+1\right)}\right]\)
\(=\frac{1}{3}\left(\frac{ab+a+b+1}{ab+a+b+1}-\frac{ab-1}{ab+a+b+1}\right)=\frac{1}{3}\left(1-\frac{0}{ab+a+b+1}\right)=\frac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}ab=1\\a^3=1\\b^3=1\end{cases}\Leftrightarrow a=b=1}\)
2) bđt \(\Leftrightarrow\)\(\left(a-b\right)^2\left(\frac{1}{a}+\frac{1}{b}\right)+\left(a+b-2\right)\left[\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\frac{7}{2}\right]\ge0\) (1)
(1) đúng do \(a+b\ge2\sqrt{ab}=2\)\(\Leftrightarrow\)\(a+b-2\ge0\)
Dấu "=" xảy ra khi a=b=1
Dấu BĐT ngược 1 chút \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)
Xét hiệu 2 vế của BĐT
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}-\frac{2}{1+ab}=\frac{1}{1+a^2}-\frac{1}{1+ab}+\frac{1}{1+b^2}-\frac{1}{1+ab}\)
\(=\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\ge0\)
=> \(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\ge\frac{2}{1+ab}\)
Dấu "=" xảy ra <=> a=b=1
1)
Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c
2)
\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)
Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
1/ Ta có: \(\frac{x^4}{1a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)
\(\Leftrightarrow1bx^4\left(a+b\right)+ay^4\left(a+b\right)=ab\left(x^4+2x^2y^2+y^4\right)\)
\(\Leftrightarrow\left(ay^2-bx^2\right)^2=0\)
\(\Rightarrow\frac{x^2}{1a}=\frac{y^2}{b}=\frac{\left(x^2+y^2\right)}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2006}}{1a^{1003}}=\frac{y^{2006}}{b^{1003}}=\frac{1}{\left(a+b\right)^{1003}}\)
\(\Rightarrow\frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{b^{1003}}=\frac{2}{\left(a+b\right)^{1003}}\)
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
Bài này chắc dùng phương pháp hạ bậc + chọn điểm rơi. :v
Lời giải:
Dự đoán dấu "=" xảy ra tại a = b = 1
Ta có: \(1+a^2\ge2a;1+b^2\ge2b\) (cô si)
Suy ra \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\le\frac{1}{2a}+\frac{1}{2b}\) (1)
Áp dụng BĐT Am-Gm (Cô si),ta có: \(ab\le\frac{a^2+b^2}{2}\)
Lại có: \(\frac{2}{1+ab}\ge\frac{2}{1+\frac{a^2+b^2}{2}}\ge\frac{2}{1+\frac{2}{2}}=1\) (2)
Ta sẽ c/m: \(\frac{1}{2a}+\frac{1}{2b}\le1\Leftrightarrow\frac{1}{a}+\frac{1}{b}\le2\)
Chứng minh tiếp đi:v,bí r:v