Set \(A=a^3+b^3+c^3\); \(B=a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\)

Apply AM-GM inequality for 2 positive numbers 

\(a^3+abc\ge2\sqrt{a^3.abc}=2a^2\sqrt{bc}\)

\(b^3+abc\ge2\sqrt{b^3.abc}=2b^2\sqrt{ac}\)

\(c^3+abc\ge2\sqrt{c^3.abc}=2c^2\sqrt{ab}\)

Thence inferred \(A+3abc\ge2B\)

Need proof: \(3abc\le A\) it mean \(3abc\le a^3+b^3+c^3\), this is always true because of the AM-GM inequality we have \(a^3+b^3+c^3\ge3\sqrt[3]{a^3.b^3.c^3}=3abc\)

Equal sign occurs when and only if a = b = c

C = 4x(x + y)(x + y + z)(x + z) + y²z²

   = 4[x(x + y + z)][(x + y)(x + z)] + y²z²

   = 4(x² + xy + xz)(x² + xy + xz + yz) + y²z²

   = 4(x² + xy + xz)² + 4yz(x² + xy + xz) + y²z²

   = [2(x² + xy + xz)]² + 2.2(x² + xy + xz).yz + (yz)²

   = [2(x² + xy + xz) + yz)]²

We have things to prove

B = 2x² + 9y² - 6xy - 6x - 12y + 2004

   = (x² + 9y² - 6xy) + (4x - 12y) + x² - 10x + 2004

   = [x² + (3y)² - 2x.3y] + 4(x - 3y) + (x² - 10x) + 2004

   = (x - 3y)² + 2.(x - 3y).2 + 4 + (x² - 2x.5 + 25) + 2004 - 4 - 25

   = (x - 3y + 2)² + (x - 5)² + 1975

We have: (x - 3y + 2)² \(\ge0\forall x;y\);  \(\left(x-5\right)^2\ge0\forall x\)

So \(\left(x-3y+2\right)^2+\left(x-5\right)^2\ge0\forall x;y\)

\(\Rightarrow\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\forall x;y\)

or \(B\ge1975\forall x;y\)

Equal sign occurs when and only if \(\left\{{}\begin{matrix}x-3y+2=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+2=3y\\x=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{3}\\x=5\end{matrix}\right.\)

x + y = 3 => (x + y)² = 9 <=> x² + y² + 2xy = 9 <=> 5 + 2xy = 9 (because x² + y² = 5)

<=> 2xy = 4 <=> xy = 2

We have: (x + y)(x² + y²) = 3.5

<=> x³ + x²y + xy² + y³ = 15

<=> x³ + y³ + xy(x + y) = 15

<=> x³ + y³ + 2.3 = 15 (because xy = 2; x + y = 3)

<=> x³ + y³ + 6 = 15

<=> x³ + y³ = 9

So x³ + y³ = 9

Applying the Cauchyschwarz inequality to the Engel form we have:

\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)

We have things to prove

x + y = 1 <=> y = 1 - x

We have: F = x² + (1 - x)² + (1 - x).x

                   = x² + 1 + x² - 2x + x - x²

                   = x² - x + 1

                   \(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Equal signs occur when and only of \(x-\dfrac{1}{2}=0\) \(\Leftrightarrow x=\dfrac{1}{2}\)

\(y=1-\dfrac{1}{2}=\dfrac{1}{2}\)

min A \(=\dfrac{1}{2}\Leftrightarrow x=y=\dfrac{1}{2}\)