Đoán xem???

ae vứt 1 ab ra nha

\(a^2+b^2+c^2+d^2+e^2\ge ab+ac+ad+ae\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\)

\(\Leftrightarrow4\left(a^2+b^2+c^2+d^2+e^2\right)\ge4a\left(b+c+d+e\right)\)

\(\Leftrightarrow\left(a^2-4ab+4b^2\right)+\left(a^2-4ac+4c^2\right)+\left(a^2-4ad+4d^2\right)+\left(a^2-4ac+4c^2\right)\ge0\)

\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\)

Bất đẳng thức đúng vậy ta có điều phải chứng minh

bài 1. ta có

\(a^2+b^2+c^2+d^2\ge ab+ac+ad\)

\(\Leftrightarrow b^2+ab+\frac{a^2}{4}+c^2+ac+\frac{a^2}{4}+d^2+ad+\frac{a^2}{4}+\frac{a^2}{4}\ge0\)

\(\Leftrightarrow\left(b+\frac{a}{2}\right)^2+\left(c+\frac{a}{2}\right)^2+\left(d+\frac{a}{2}\right)^2+\frac{a^2}{4}\ge0\) luôn đúng

Bài 2

ta có \(\frac{a^5}{b^5}+1+1+1+1\ge\frac{5.a}{b}\) (bất đẳng thức cauchy)

Tương tự ta có \(\frac{b^5}{c^5}+4\ge\frac{5b}{c};\frac{c^5}{a^5}+4\ge\frac{5c}{a}\)

\(\Rightarrow\frac{a^5}{b^5}+\frac{b^5}{c^5}+\frac{c^5}{a^5}\ge5\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-12\)

Mà dễ dàng chứng minh \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\)

Nên ta có \(\Rightarrow\frac{a^5}{b^5}+\frac{b^5}{c^5}+\frac{c^5}{a^5}\ge5\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-12\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

bài 1 : \(^{a^2+B^2+C^2+D^2}\)>hoặc =ab+ac+ad 

\(^{a^2+b^2+c^2}\)- ab-ac-ad>hoặc = 0

\((\frac{1}{4}^{a^2-ab+b^2})+(\frac{1}{4}^{a^2-ac+c^2})+(\frac{1}{4}^{a^2-ad+d^2})\)>hoặc =0

\((\frac{1}{2}a-b)^2+(\frac{1}{2}a-c)^2+(\frac{1}{2}a-d)^2>=0\)

Vì \((\frac{1}{2}a-b)^2>=0\)với mọi \(A,b\varepsilon n\)

=> đpcm tự kết luận

BĐT\(\Leftrightarrow3a^2+3b^2+3c^2+3d^2+6\left(ab+bc+cd+da+bd+ca\right)\ge8\left(ab+bc+cd+da+bd+ca\right)\)

\(\Leftrightarrow3a^2+3b^2+3c^2+3d^2-2\left(ab+bc+cd+da+bd+ca\right)\ge0\) (*)

Ta có: \(a^2+b^2\ge2ab;b^2+c^2\ge2bc;c^2+d^2\ge2cd\)

\(d^2+a^2\ge2da;b^2+d^2\ge2bd;c^2+a^2\ge2ca\)

Cộng theo vế các BĐT trên suy ra \(3a^2+3b^2+3c^2+3d^2\ge2\left(ab+bc+cd+da+bd+ca\right)\)

Do vậy BĐT (*) đúng hay ta có đpcm.

P/s: EM còn khá dốt BĐT,mong được các anh chị chỉ bảo cho ạ!

Cần cù bù thông minh ^^

\(BDT\Leftrightarrow\frac{1}{9}\left(-3a+b+c+d\right)^2+\frac{2}{9}\left(2b-c-d\right)^2+\frac{2}{3}\left(c-d\right)^2\ge0\)

Hihi mình phân tích hơi nham nhở thông cảm nha :(

Thử cách này xem sao:

BĐT \(\Leftrightarrow\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2+\left(a-d\right)^2+\left(b-d\right)^2+\left(c-d\right)^2}{3}\ge0\) (đúng)

Vậy ta có đpcm.

Ta có : \(\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{2}{3}\)

\(\Leftrightarrow3\left(a+b+c+d\right)^2\ge8\left(ab+ac+ad+bc+bd+cd\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2+d^2\right)+6\left(ab+ac+ad+bc+bd+cd\right)\ge8\left(ab+ac+ad+bc+bd+cd\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2+d^2\right)-2\left(ab+ac+ad+bc+bd+cd\right)\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(a^2-2ad+d^2\right)+\left(b^2-2bc+c^2\right)+\left(b^2-2bd+d^2\right)+\left(c^2-2cd+d^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(c-d\right)^2\ge0\) (luôn đúng)

Vậy bđt ban đầu được chứng minh

\(a^2+b^2+c^2+d^2+e^2\ge ab+ac+ad+ae\)

Ta có :

\(a^2+b^2+c^2+d^2+e^2\)

\(=\left(\dfrac{a^2}{4}+b^2\right)+\left(\dfrac{a^2}{4}+c^2\right)+\left(\dfrac{a^2}{4}+d^2\right)+\left(\dfrac{a^2}{4}+e^2\right)\)

Ta lại có :

\(\left(\dfrac{a}{2}-b\right)^2\ge0\Leftrightarrow\) \(\dfrac{a^2}{4}-ab+b^2\ge0\) \(\dfrac{\Rightarrow a^2}{4}+b^2\ge ab\)

Tương tự :

\(\dfrac{a^2}{4}+c^2\ge ac\)

\(\dfrac{a^2}{4}+d^2\ge ad\)

\(\dfrac{a^2}{4}+e^2\ge ae\)

\(\Rightarrow\left(\dfrac{a^2}{4}+b^2\right)+\left(\dfrac{a^2}{4}+c^2\right)+\left(\dfrac{a^2}{4}+d^2\right)+\left(\dfrac{a^2}{4}+e^2\right)\ge ab+ac+ad+ae\)

\(\Rightarrow a^2+b^2+c^2+d^2+e^2\ge ab+ac+ad+ae\)

a.

\(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)

\(\Leftrightarrow2a^4+2b^4\ge a^4+ab^3+a^3b+b^4\)

\(\Leftrightarrow a^4+b^4\ge ab^3+a^3b\)

\(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)

\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)\ge0\)

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

Mà \(a^2+ab+b^2=\left(a^2+2\cdot a\cdot\dfrac{1}{2}b+\dfrac{b^2}{4}\right)+\dfrac{3b^2}{4}=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\)

Suy ra (*) đúng => đpcm

Dấu "=" xảy ra khi a = b

b.

\(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)

\(\Leftrightarrow3a^4+3b^4+3c^4\ge a^4+ab^3+ac^3+a^3b+b^4+bc^3+a^3c+b^3c+c^4\)

\(\Leftrightarrow2a^4+2b^4+2c^4\ge ab^3+a^3b+b^3c+bc^3+ca^3+c^3a\)

\(\Leftrightarrow\left(a^4+b^4\right)+\left(b^4+c^4\right)+\left(c^4+a^4\right)\ge\left(a^3b+ab^3\right)+\left(b^3c+bc^3\right)+\left(c^3a+ca^3\right)\)

Theo câu a. thì điều này đúng

Dấu "=" khi a=b=c

c. Theo bất đẳng thức Cauchy-Schwarz, ta có:

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\dfrac{\left(a+b+c\right)^2}{b+c+a}=a+b+c\left(dpcm\right)\)

Dấu "=" xảy ra khi \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Leftrightarrow a=b=c\)