Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)

Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3

\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)

\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)

\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)

ta sẽ giết ngươi kí tên dép đờ kiu lờ

thử bài bất :D 

Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)

Hoàn toàn tương tự: 

\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)

\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)

Cộng (*),(**),(***) vế theo vế ta được:

\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)

Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )

Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi a=b=c=1

1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D 

Bài 6 . Áp dụng BĐT Cauchy , ta có :

a² + b² ≥ 2ab ( a > 0 ; b > 0)

⇔ ( a + b)² ≥ 4ab

⇔ \(\dfrac{\left(a+b\right)^2}{4}\)≥ ab

⇔ \(\dfrac{a+b}{4}\) ≥ \(\dfrac{ab}{a+b}\) ( 1 )

CMTT , ta cũng được : \(\dfrac{b+c}{4}\) ≥ \(\dfrac{bc}{b+c}\) ( 2) ; \(\dfrac{a+c}{4}\) ≥ \(\dfrac{ac}{a+c}\)( 3)

Cộng từng vế của ( 1 ; 2 ; 3 ) , Ta có :

\(\dfrac{a+b}{4}\) + \(\dfrac{b+c}{4}\) + \(\dfrac{a+c}{4}\) ≥ \(\dfrac{ab}{a+b}\) + \(\dfrac{bc}{b+c}\) + \(\dfrac{ac}{a+c}\)

⇔ \(\dfrac{a+b+c}{2}\) ≥ \(\dfrac{ab}{a+b}\) + \(\dfrac{bc}{b+c}\) + \(\dfrac{ac}{a+c}\)

Bài 4.

Áp dụng BĐT Cauchy cho các số dương a , b, c , ta có :

\(1+\dfrac{a}{b}\) ≥ \(2\sqrt{\dfrac{a}{b}}\) ( a > 0 ; b > 0) ( 1)

\(1+\dfrac{b}{c}\) ≥ \(2\sqrt{\dfrac{b}{c}}\) ( b > 0 ; c > 0) ( 2)

\(1+\dfrac{c}{a}\) ≥ \(2\sqrt{\dfrac{c}{a}}\) ( a > 0 ; c > 0) ( 3)

Nhân từng vế của ( 1 ; 2 ; 3) , ta được :

\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\) ≥ \(8\sqrt{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}}=8\)

@Giáo Viên Hoc24.vn

@Akai Haruma

Tính được A + B = a - c - 4; và C - D = a - c - 4, từ đó suy ra ĐPCM

Lm dc hết chưa bn ơi.

câu 2

a^4 + b^4 + c^4 + d^4 = 4abcd

<=> \(a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4+2a^2b^2-4abcd+2b^2d^2=0\)

<=> \(\left(a^2-b^2\right)^2+\left(c^2-d^2\right)^2+2\left(ab-cd\right)^2=0\)

<=> \(\left\{{}\begin{matrix}a^2=b^2\\c^2=d^2\\ab=cd\end{matrix}\right.\Leftrightarrow a=b=c=d\)

Ta có \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)[\left(a+b\right)^2-c\left(a+b\right)+c^2]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-bc-ca+c^2-3ba\right)\)

hay \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Ta lại có \(a^3+b^3+c^3=3abc\) khi \(\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

khi a,b,c dương thì a+b+c ko thể =0 nên a=b=c

F(0)=d⇒d⋮5F(0)=d⇒d⋮5

F(1)=a+b+c+d⋮5⇒a+b+c⋮5F(1)=a+b+c+d⋮5⇒a+b+c⋮5

F(−1)=−a+b−c+d⋮5⇒−a+b−c⋮5F(−1)=−a+b−c+d⋮5⇒−a+b−c⋮5

⇒(a+b+c)+(−a+b−c)⋮5⇒(a+b+c)+(−a+b−c)⋮5

⇒2b⋮5⇒b⋮5⇒2b⋮5⇒b⋮5

⇒a+c⋮5