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

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

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 :

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

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

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

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

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

cảm ơn bạn

BĐT cần c/m tương đương:

\(2\left(a^3+b^3+c^3+d^3\right)\ge2+\dfrac{3}{2}\sqrt{4+2\left(ab+ac+ad+bc+bd+cd\right)}\)

\(\Leftrightarrow2\left(a^3+b^3+c^3+d^3\right)\ge2+\dfrac{3}{2}\sqrt{\left(a+b+c+d\right)^2}\)

\(\Leftrightarrow2\left(a^3+b^3+c^3+d^3\right)\ge2+\dfrac{3}{2}\left(a+b+c+d\right)\)

\(\Leftrightarrow4\left(a^3+b^3+c^3+d^3\right)\ge4+3\left(a+b+c+d\right)\)

Dễ dàng chứng minh điều này bằng AM-GM:

\(a^3+a^3+1+b^3+b^3+1+c^3+c^3+1+d^3+d^3+1\ge3a^2+3b^2+3c^2+3d^2\)

\(\Rightarrow2\left(a^3+b^3+c^3+d^3\right)+4\ge12\)

\(\Rightarrow a^3+b^3+c^3+d^3\ge4\) (1)

Lại có:

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

\(\Rightarrow a+b+c+d\le4\) (2)

(1);(2) \(\Rightarrow4\left(a^3+b^3+c^3+d^3\right)\ge16\ge4+3.4\ge4+3\left(a+b+c+d\right)\) (đpcm)

1a) 8x + 15 - 3x = -400                          b) -32x + 12x - 5x = 900

=> 5x = -400 - 15                                   => -25x = 900

=> 5x = -415                                          => x = 900 : (-25)

=> x = -415 : 5                                      => x = -36

=> x = -83

c) 3(x - 1) - (x - 5) = -18

=> 3x - 3 - x + 5 = -18

=> 2x + 2 = -18

=> 2x = -18 - 2

=> 2x = -20

=> x = -10

d,e tự làm

a) \(8x+15-3x=-400\)

\(\Leftrightarrow8x-3x=-400-15\)

\(\Leftrightarrow5x=-415\)

\(\Leftrightarrow x=-415\div5\)

\(\Leftrightarrow x=-83\)

b) \(-32x+12x-5x=900\)

\(\Leftrightarrow-25x=900\)

\(\Leftrightarrow x=900\div\left(-25\right)\)

\(\Leftrightarrow x=-36\)

Bài 2:ĐKXĐ: x<>0

\(2\left(x^2+\frac{1}{x^2}\right)-x-\frac{1}{x}=6\)

=>\(2\left\lbrack\left(x+\frac{1}{x}\right)^2-2\cdot x\cdot\frac{1}{x}\right\rbrack-\left(x+\frac{1}{x}\right)=6\)

=>\(2\left(x+\frac{1}{x}\right)^2-4-\left(x+\frac{1}{x}\right)-6=0\)

=>\(2\left(x+\frac{1}{x}\right)^2-\left(x+\frac{1}{x}\right)-10=0\)

=>\(2\left(x+\frac{1}{x}\right)^2-5\left(x+\frac{1}{x}\right)+4\left(x+\frac{1}{x}\right)-10=0\)

=>\(\left(x+\frac{1}{x}\right)\left\lbrack2\left(x+\frac{1}{x}\right)-5\right\rbrack+2\cdot\left\lbrack2\left(x+\frac{1}{x}\right)-5\right\rbrack=0\)

=>\(\left(x+\frac{1}{x}+2\right)\left(2x+\frac{2}{x}-5\right)=0\)

=>\(\frac{x^2+2x+1}{x}\cdot\frac{2x^2-5x+2}{x}=0\)

=>\(\left(x+1\right)^2\left(2x^2-5x+2\right)=0\)

=>\(\left(x+1\right)^2\cdot\left(x-2\right)\left(2x-1\right)=0\)

=>\(\left[\begin{array}{l}x+1=0\\ x-2=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\left(nhận\right)\\ x=2\left(nhận\right)\\ x=\frac12\left(nhận\right)\end{array}\right.\)

Bài 1:

a: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=6^2+8^2=100=10^2\)

=>BC=10

Xét ΔABC có AD là phân giác

nên \(\frac{BD}{AB}=\frac{CD}{AC}\)

=>\(\frac{BD}{6}=\frac{CD}{8}\)

mà BD+CD=BC=10

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{BD}{6}=\frac{CD}{8}=\frac{BD+CD}{6+8}=\frac{10}{14}=\frac57\)

=>\(\begin{cases}BD=6\cdot\frac57=\frac{30}{7}\\ CD=8\cdot\frac57=\frac{40}{7}\end{cases}\)

b: Xét ΔCEB có AD//EB

nên \(\frac{CA}{AE}=\frac{CD}{DB}\)

=>\(\frac{8}{AE}=\frac43=\frac86\)

=>AE=6(cm)

c: Ta có: DI⊥AC

AB⊥CA

Do đó: DI//AB

Xét ΔCAB có DI//AB

nên \(\frac{CI}{IA}=\frac{CD}{DB}\)

=>\(\frac{CI}{IA}=\frac{CA}{AB}\)