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

j vậy bẹn, đây là sinh lớp 7 mak :v ?

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

Đề hay thật sự, cho x,y,z nhưng chứng minh a,b,c :v

1. (a+b)^2 ≥ 4ab

<=> a²+2ab+b²≥ 4ab

<=> a²+2ab+b²-4ab≥ 0

<=> a²-2ab+b²≥ 0

<=> (a-b)^2 ≥ 0 ( luôn đúng )

2. a^2 + b^2 + c^2 ≥ ab + bc + ca

<=> 2a^2 + 2b^2 + 2c^2 ≥ 2ab + 2bc + 2ca

<=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca ≥ 0

<=> (a^2- 2ab+b^2) + (b^2-2bc+c^2) + (c^2-2ca+a^2) ≥ 0

<=> (a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 0 ( luôn đúng)

4. Tương tự 3

\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2\ge5\sqrt[5]{\dfrac{a^{20}b^2}{b^{12}}}=5.\dfrac{a^4}{b^2}\)

\(\Rightarrow4.\dfrac{a^5}{b^3}+b^2\ge5.\dfrac{a^4}{b^2}\)

Tương tự: \(4.\dfrac{b^5}{c^3}+c^2\ge5\dfrac{b^4}{c^2};4\dfrac{c^5}{a^3}+a^2\ge5.\dfrac{c^4}{a^2}\)

\(\Rightarrow4\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)

Lại có: \(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2+b^2\ge5a^2\)

\(\Rightarrow2.\dfrac{a^5}{b^3}+3b^2\ge5a^2\), tương tự: \(2.\dfrac{b^5}{c^3}+3c^2\ge5b^2;2\dfrac{c^5}{a^3}+3a^2\ge5c^2\)

\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)

\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}+4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5.\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)

\(\Rightarrow dpcm\)

giả sử \(a>b>c>0\) thì ta có :

\(\dfrac{a^4}{b^2}\left(\dfrac{a}{b}-1\right)+\dfrac{b^4}{c^2}\left(\dfrac{b}{c}-1\right)+\dfrac{c^4}{a^2}\left(\dfrac{c}{a}-1\right)\ge\dfrac{2a^2b}{c}+\dfrac{c^5}{a^3}-\dfrac{c^4}{a^2}\)

\(\ge\dfrac{2c^4b}{a}-\dfrac{c^4}{a^2}=\dfrac{c^4}{a}\left(2b-\dfrac{1}{a}\right)>0\)

làm tương tự cho trường hợp \(c>b>a>0\) ; \(b>a>c\) và \(b>c>a\)

\(\Rightarrow\left(đpcm\right)\)

mấy câu cậu câu đăng khác bn làm tương tự nha . nếu bn lm không được thì có j mk lm luôn cho còn h mk bạn rồi :(