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\(\dfrac{a^3+b^3}{2}\ge\left(\dfrac{a+b}{3}\right)^3\)

<=> \(8.\dfrac{a^3+b^3}{2}\ge a^3+3a^2b+3ab^2+b^3\)

<=> \(3\left(a^3+b^3\right)\ge3a^2b+3ab^2\)

<=> \(a^3+b^3-a^2b-ab^3\ge0\)

<=> \(\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)

<=> \(\left(a+b\right)\left(a-b\right)^2\ge0\)  (Right)

So ...... 

The title must be: a + b = 2

(1st condition): a \(\le\) 0, b \(\ge\) 0

=> \(a^4\ge0;b^4\ge0\)

We have: a + b = 2 with \(a^4;b^4\in Z\)+

=> \(a^4+b^4\ge a+b\ge2\) (proved)

(2nd condition): a \(\ge\) 0; b \(\le\) 2: Do exactly as the 1st condition (proved)

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

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 ............

who fast i

We have:\(x:y:z=a:b:c\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=x+y+z\)

\(\Rightarrow\left(x+y+z\right)^2=\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\)

We have 

\(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{b}{c}\Leftrightarrow\dfrac{10a+b}{10b+c}=\dfrac{b}{c}=\dfrac{10a+b-b}{10b+c-c}=\dfrac{10a}{10b}=\dfrac{a}{b}\)

=> \(\dfrac{a}{b}=\dfrac{b}{c}\)

64 squares

\(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2+2abc+1\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2+4abc+2\ge2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge-4abc-2\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge-\left(abc+2\right)\) (True because -(abc+2) \(\le0\) because a,b,c > 0) 

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

Oh, sorry, the last row must be \(\ge-\left(2abc+2\right)\) not \(\ge-\left(abc+2\right)\)