xét hiệu \(\frac{1}{1+a^2}-\frac{1}{1+ab}+\frac{1}{1+b^2}-\frac{1}{1+ab}\)

quy đồng làm nốt nha                                

\(\frac{1}{\left(1+a^2\right)}+\frac{1}{\left(1+b^2\right)}>=\frac{2}{\left(1+ab\right)}\)

\(\Leftrightarrow\frac{1}{\left(1+a^2\right)}+\frac{1}{\left(1+b^2\right)}-\frac{2}{\left(1+ab\right)}>=0\)

\(\Leftrightarrow\left[\frac{1}{\left(1+a^2\right)}-\frac{1}{\left(1+ab\right)}\right]+\left[\frac{1}{\left(1+b^2\right)}-\frac{1}{\left(1+ab\right)}\right]>=0\)

\(\Leftrightarrow\left[\frac{a\left(b-c\right)}{\left(1+a^2\right)\left(1+ab\right)}\right]+\left[\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\right]>=0\)

\(\frac{\left[a\left(b-a\right)\left(1+b^2\right)-b\left(b-a\right)\left(1+a^2\right)\right]}{\left[\left(1+a^2\right)\left(1+b^2\right)\left(1+b^2\right)\left(1+ab\right)^2\right]}>=0\)

\(\frac{\left[\left(b-a\right)\left(a+ab^2-b+ba^2\right)\right]}{\left[\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)^2\right]}>=0\)

\(\frac{\left[\left(b-a\right)\left[\left(a-b\right)+ab\left(b-a\right)\right]\right]}{\left[\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)^2\right]}>=0\)

\(\frac{\left[\left(b-a\right)^2\left(ab-1\right)\right]}{\left[\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)^2\right]}>=0\left(1\right)\)

Mẫu số luôn lớn hơn 1 

\(\left(b-a\right)^2>=0\)  voi moi a,b

Vì a,b >=1 nên ( ab-1) > = 0

​Nên (1)  dụng

Tu "dung"doi thanh dung

a/ Đề sai, đề đúng phải là \(p=\frac{a+b+c}{2}\)

b/ \(\Leftrightarrow\frac{2}{2+a^2b}+\frac{2}{2+b^2c}+\frac{2}{2+c^2a}\ge2\)

\(VT=1-\frac{a^2b}{1+1+a^2b}+1-\frac{b^2c}{1+1+b^2c}+1-\frac{c^2a}{1+1+c^2a}\)

\(VT\ge3-\left(\frac{a^2b}{3\sqrt[3]{a^2b}}+\frac{b^2c}{3\sqrt[3]{b^2c}}+\frac{c^2a}{3\sqrt[3]{c^2a}}\right)\)

\(VT\ge3-\frac{1}{9}\left(3\sqrt[3]{a^2.ab.ab}+3\sqrt[3]{b^2.bc.bc}+3\sqrt[3]{c^2.ca.ca}\right)\)

\(VT\ge3-\frac{1}{9}\left(a^2+2ab+b^2+2bc+c^2+2ca\right)\)

\(VT\ge3-\frac{1}{9}\left(a+b+c\right)^2=2\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Bạn cần biết  \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)  (nếu bạn chưa biết thì xét hiệu) 

Ta có: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\)

\(\ge\frac{4}{1+a^2+1+b^2}\)

\(=\frac{4}{a^2+b^2+2}\)

\(\ge\frac{4}{2ab+2}=\frac{2}{ab+1}\)

Dấu "=" xảy ra khi \(a=b\)

\(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=b^2+2bc+c^2\\b^2=a^2+2ac+c^2\\c^2=a^2+2ab+b^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b^2+c^2-a^2=-2bc\\a^2+c^2-b^2=-2ac\\a^2+b^2-c^2=-2ab\end{matrix}\right.\Rightarrow P=\frac{1}{-2bc}+\frac{1}{-2ac}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=0\)

a) \(P=\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+b^2-c^2}+\frac{1}{a^2+c^2-b^2}\) ( Sửa đề )

\(P=\frac{1}{\left(b+c\right)^2-2ab-a^2}+\frac{1}{\left(a+b\right)^2-2ab-c^2}+\frac{1}{\left(a+c\right)^2-2ac-b^2}\)

Vì a + b + c = 0

Nên a + b = -c

=> ( a + b )² = (-c)² = c²

Tương tự: ( b + c )² = a² và ( a + c )² = b²

\(\Rightarrow P=\frac{1}{a^2-2bc-a^2}+\frac{1}{c^2-2ab-c^2}+\frac{1}{b^2-2ac-b^2}\)

\(P=\frac{1}{-2bc}+\frac{1}{-2ab}+\frac{1}{-2ac}\)

\(P=\frac{a+b+c}{-2abc}=\frac{0}{-2abc}=0\)

\(xét:\frac{1}{a^2+1}+\frac{1}{b^2+1}-\frac{2}{1+ab}=\left(\frac{1}{a^2+1}-\frac{1}{1+ab}\right)+\left(\frac{1}{b^2+1}-\frac{1}{1+ab}\right)=\frac{1+ab-a^2-1}{\left(a^2+1\right)\left(1+ab\right)}+\frac{1+ab-1-b^2}{\left(b^2+1\right)\left(1+ab\right)}=\frac{a\left(b-a\right)}{\left(a^2+1\right)\left(1+ab\right)}+\frac{b\left(a-b\right)}{\left(b^2+1\right)\left(1+ab\right)}=\left(a-b\right)\left(\frac{b}{\left(b^2+1\right)\left(1+ab\right)}-\frac{a}{\left(a^2+1\right)\left(1+ab\right)}\right)=\left(a-b\right)\left(\frac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(ab+1\right)\left(b^2+1\right)}\right)=\left(a-b\right)\left(\frac{\left(ab-1\right)\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\) \(\left(a-b\right)^2\frac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\left(do:a\ge1;b\ge1\right)\Rightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\left(a\ge1;b\ge1\right)\)

1. Ta có : \(\frac{a}{a+b+c+d}< \frac{a}{a+b+c}< \frac{a+d}{a+b+c+d}\)

          \(\frac{b}{a+b+c+d}< \frac{b}{b+c+d}< \frac{a+b}{a+b+c+d}\)

          \(\frac{c}{a+b+c+d}< \frac{c}{a+c+d}< \frac{b+c}{a+b+c+d}\)

         \(\frac{d}{a+b+c+d}< \frac{d}{a+b+d}< \frac{c+d}{a+b+c+d}\)

Cộng vế theo vế ta được :

\(1< \frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2\)             ( đpcm )

2. Áp dụng bất đẳng thức Cô - si cho 2 số ko âm b-1 và 1 ta có :

\(\sqrt{\left(b-1\right)\cdot1}\le\frac{\left(b-1\right)+1}{2}=\frac{b}{2}\)

Dấu "=" xảy ra <=> b - 1 = 1    <=> b = 2

\(\Rightarrow a\sqrt{b-1}=a\sqrt{\left(b-1\right)\cdot1}\le a\cdot\frac{b}{2}=\frac{ab}{2}\)

Tương tự ta có : \(b\sqrt{a-1}\le\frac{ab}{2}\) Dấu "=" xảy ra <=> a = 2

Do đó : \(a\sqrt{b-1}+b\sqrt{a-1}\le\frac{ab}{2}+\frac{ab}{2}=ab\)

Dấu "=" xảy ra <=> a = b = 2

\(\Leftrightarrow\left(2+a^2+b^2\right)\left(1+ab\right)\ge2\left(1+a^2\right)\left(1+b^2\right)\)

\(\Leftrightarrow2+2ab+a^2+b^2+ab\left(a^2+b^2\right)\ge2+2a^2+2b^2+2a^2b^2\)

\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\) (luôn đúng với mọi \(a\ge1;b\ge1\))

Cách khác:

\(\Leftrightarrow\left(\frac{1}{1+a^2}-\frac{1}{1+ab}\right)+\left(\frac{1}{1+b^2}-\frac{1}{1+ab}\right)\ge0\)

\(\Leftrightarrow\frac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)\left[b\left(1+a^2\right)-a\left(1+b^2\right)\right]}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\) (luôn đúng).