Apply inequality cauchy, we have:

⎧⎩⎨⎪⎪⎪⎪(b+c−a)+(a+c−b)≥2(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−√(a+c−b)+(a+b−c)≥2(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−√(a+b−c)+(b+c−a)≥2(a+b−c)(b+c−a)−−−−−−−−−−−−−−−−−√

⇒⎧⎩⎨⎪⎪⎪⎪2c≥2(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−√2a≥2(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−√2b≥2(a+b−c)(b+c−a)−−−−−−−−−−−−−−−−−√

⇒⎧⎩⎨⎪⎪c2≥(b+c−a)(a+c−b)a2≥(a+c−b)(a+b−c)b2≥(a+b−c)(b+c−a)

⇒(abc)2≥[(b+c−a)(a+c−b)(a+b−c)]2

⇒abc≥(b+c−a)(a+c−b)(a+b−c)

The title must be: a + b = 2

(1st condition): a ≤

 0, b ≥

 0

=> a4≥0;b4≥0

We have: a + b = 2 with a4;b4∈Z

+

=> a4+b4≥a+b≥2

 (proved)

(2nd condition): a ≥

 0; b ≤

 2: Do exactly as the 1st condition (proved)

So ............

We have: a2+b2+c2≥ab+bc+ca

[a2+b2+c2≥ab+bc+ca⇔2(a2+b2+c2)≥2(ab+ac+ca)⇔(a−b)2+(b−c)2

≥0

 (proved)]

=> a2+b2+c2+2ab+2bc+2ca≥ab+bc+ca

=> (a+b+c)2≥3(ab+ac+ca)

=> (a+b+c)3≥3(ab+ac+ca)

 (proved)

a4+b4≥a3b+ab3

<=> a4+b4−a3b−ab3≥0

<=> a3(a−b)−b3(a−b)≥0

<=> (a3−b3)(a−b)≥0

<=> (a−b)2(a2+ab+b2)≥0

Cause ⎧⎩⎨⎪⎪⎪⎪(a−b)2≥0∀a,b∈Ra2+ab+b2=(a+b2)2+3b24≥0

You have to have a condition : a,b>0

a) (a+b)2≥4ab

<=> a2+2ab+b2≥4ab

<=>  a2−2ab+b2≥0

<=>  (a−b)2≥0

  (It's true)

=> (a+b)2≥4ab

b) Applying Cauchy's inequality , we have

a+b≥2ab−−√

Equation occur <=> a = b

We have:

a2+b2+c2≥ab+bc+ca

Prove that it: 

This inequality <=> 2a2+2b2+2c2≥2ab+2bc+2ca

<=> 2a2+2b2+2c2−2ab−2bc−2ca≥0

<=> (a−b)2+(b−c)2+(c−a)2≥0

 (true) 

So a2 + b2 + c2 ≥

 ab + bc + ca

Apply this for the topic, we have:

a4+b4+c4≥(ab)2+(bc)2+(ca)2

(ab)2+(bc)2+(ca)2≥ab2c+a2bc+abc2=abc(a+b+c)

So a4+b4+c4≥abc(a+b+c)

a+b+c = 0

=> (a+b+c)2 = 0

=> a2 + b2 + c2 + 2ab + 2bc + 2ca = 0

We have: a2 + b2 + c2 ≥0

⇒2(ab+bc+ca)≤0⇔ab+bc+ca≤0

8(a3+b3+c3)≥(a+b)3+(b+c)3+(c+a)3

⇔8(a3+b3+c3)≥2(a3+b3+c3)+3ab(a+b)+3bc(b+c)+3ca(c+a)

⇔

6(a3+b3+c3)≥3[ab(a+b)+bc(b+c)+ca(c+a)]

⇔

 2a3+2b3+2c3≥ab(a+b)+bc(b+c)+ca(c+a)

We have: a3 + b3 ≥

 a2b + ab2

Prove that the same way as the way of Kaya Renger to prove that: a4+b4≥a3b+ab3

So ⎧⎩⎨⎪⎪a3+b3≥ab(a+b)b3+c3≥bc(b+c)c3+a3≥ca(c+a)

So true. 

=> 8(a3+b3+c3)≥(a+b)3+(b+c)3+(c+a)3

Applying Cauchy's inequality , we have 

(a4+1)+(b4+1)≥2a2+2b2=2(a2+b2)≥2.2ab=4ab

Equation occur <=> a = b = ±1

Apply inequality Cauchy, we have:

a+b≥2ab−−√

⇒{a4+b4≥2a4b4−−−−√c4+d4≥2c4d4−−−−√

⇔a4+b4+c4+d4≥2a2b2+2c2d2=2[(ab)2+(cd)2]

But (ab)2+(cd)2≥2(ab)2⋅(cd)2−−−−−−−−−√=2abcd

⇔a4+b4+c4+d4≥2⋅2abcd=4abcd

e have :

=x+289+1+x+586+1>x+883+1+x+1180+1

=x+9189+x+9186>x+9183+x+9180

=x+9189+x+9186−x+9183−x+9180>0

=(x+91)(189+186−183−180)>0

To 189+186−183−180

different 0

=x+91>0

=x<−91

We have :

To

different 0

We have:

1n2+(n+1)2=12n2+2n+1=12⎛⎜ ⎜ ⎜⎝1n2+n+12⎞⎟ ⎟ ⎟⎠<12(1n2+n)=12(1n−1n+1)

Apply, we have:

15+113+125+...+1n2+(n+1)2<12(1−12)+12(13−14)+12(14−15)+...+12(1n−1n+1)=12(1−1n+1)<12

Let me answer you uncle : DO NOT KNOW

a , x2(x−1)+(x−1)=0

(x−1)(x2+1)=0

x=1

b , (x−2)(x2−16)=0

(x−2)(x−4)(x+4)=0

x=2,x=4,x=4

tan(a)=12⇔a=26,5605118

⇒tan(90−2a)=tan(90−2⋅26,56505118)=34

a , x8+14x4+1=(x4+1)2+12x4

Add and subtract 4x2(x4+1)

we have :

(x4+1)2+4x2(x4+1)+4x4−4x2(x4+1)+8x4

=(x4+1+2x2)2−4x2(x4+1−2x2)

=(x4+2x2+1)2−4x2(x2−1)2

=(x4+2x2+1)2−(2x3−2x)2

=(x4+2x3+2x2−2x+1)(x4−2x3+2x2+2x+1)

b , x8+98x4+1=(x4+1)2+96x4

=(x4+1)2+16x2(x4+1)+64x4−16x2(x4+1)+32x4

=(x4+1+8x2)2−16x2(x4+1−2x2)

=(x4+8x2+1)2−16x2(x2−1)2

=(x4+8x2+1)2−(4x3−4x)2

=(x4+4x3+8x2−4x+1)(x4−4x3+8x2+4x+1)

We have : 2x−1x−1+1=1x−1

2x−1x−1+x−1x−1=1x−1

2x−1+x−1x−1=1x−1

3x−2x−1−1x−1=0

3x−2−1x−1=0

3x−3x−1=0

3(x−1)x−1=0

3=0

So the equation has no solution .

We have : x2−5xx−5=5

x(x−5)x−5=5

x=5

So the experiment of the above equation is S={5}

We have: 13=1;23=8;...;63=21613=1;23=8;...;63=216

=> # =73=343