\(1)\)
\(A=a\left(a^2+2b\right)+b\left(b^2-a\right)=a^3+2ab+b^3-ab=a^3+b^3+ab\)
\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+ab=a^2+b^2\ge\frac{\left(a+b\right)^2}{1+1}=\frac{1}{2}\) ( Cauchy-Schwarz dạng Engel )
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=\frac{1}{2}\)
\(2)\)
\(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}=\frac{1}{\frac{a+b+c}{2}-a}+\frac{1}{\frac{a+b+c}{2}-b}+\frac{1}{\frac{a+b+c}{2}-c}\)
\(=2\left(\frac{1}{-a+b+c}+\frac{1}{a-b+c}+\frac{1}{a+b-c}\right)\)
Có : \(\hept{\begin{cases}b-a< c\\c-b< a\\a-c< b\end{cases}}\)
\(2\left(\frac{1}{-a+b+c}+\frac{1}{a-b+c}+\frac{1}{a+b-c}\right)>2\left(\frac{1}{2c}+\frac{1}{2a}+\frac{1}{2b}\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) ???
1. A = a(a2 + 2b) + b(b2 - a)
A = a3 + 2ab + b3 - ab
A = a3 + ab + b3
A = ( a + b ) ( a2 - ab + b2 ) + ab
A = a2 + b2
Mà ( a - b )2 \(\ge\)0 với mọi a,b
\(\Rightarrow\)a2 + b2 \(\ge\)2ab \(\Rightarrow\)2 . ( a2 + b2 ) \(\ge\)( a + b )2 = 1 \(\Rightarrow\)( a2 + b2 ) \(\ge\)\(\frac{1}{2}\)
\(\Rightarrow\)A \(\ge\)\(\frac{1}{2}\) . Dấu " = " xảy ra \(\Leftrightarrow\)a = b \(\frac{1}{2}\)
2) vì a,b,c là 3 cạnh của 1 tam giác nên a,b,c > 0 ; p - a > 0 ; p - b > 0 ; p - c > 0
Áp dụng BĐT : \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\). Dấu " = " xảy ra \(\Leftrightarrow\)x = y
Ta có : \(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{2p-a-b}=\frac{4}{c}\)
Tương tự : \(\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{2p-b-c}=\frac{4}{a};\frac{1}{p-c}+\frac{1}{p-a}\ge\frac{4}{2p-c-a}=\frac{4}{b}\)
Cộng từng vế 3 BĐT, ta được :
\(2.\left(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\right)\ge4.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c
Giới thiệu luon bđt mới nhé
Với mọi b dương và a tuỳ ý ta luôn có :
\(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\frac{a_3^2}{b_3}+...+\frac{a_n^2}{b_n}\ge\frac{\left(a_1+a_2+a_3+...+a_n\right)^2}{b_1+b_2+b_3+...+b_n}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}=...=\frac{a_n}{b_n}\)
\(2)\)
\(\frac{1}{p-a}+\frac{1}{p-b}>\frac{4}{2p-a-b}=\frac{4}{c}\)
\(\frac{1}{p-b}+\frac{1}{p-c}>\frac{4}{2p-b-c}=\frac{4}{a}\)
\(\frac{1}{p-c}+\frac{1}{p-a}>\frac{4}{2p-c-a}=\frac{4}{b}\)
\(VT>\frac{\frac{4}{a}+\frac{4}{b}+\frac{4}{c}}{2}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\) hay tam giác đó là tam giác đều
