$b=0$ is trivial, $b\neq 0$ devide both sides $b^4$ and denote $t= \frac{a}{b}$ and above ineq turns out to be $2(t^4+t^2+1)\geq 3(t^3+t) <=> (t-1)^2(2t^2+t+2)\geq 0$

This can be easily written as $(a-b)^{4}+a^{4}+b^{4}+a^{3}b+ab^{3}\geq{4a^{2}b^{2}}$ which follows by AM-GM.

Different: Let $a^2 + b^2 = m$ and $ab = n$; then we must show $$2(m^2 - n^2) \geq 3mn \Longleftrightarrow (2m + n)(m - 2n) \geq 0$$ Note that $2m + n = 2a^2 + ab + 2b^2 > 0$, and $m - 2n = a^2 - 2ab + b^2 \geq 0$ by AM-GM, so we are done.

Generalization Let $a,b,x,y$ be real numbers with ${0<y\leq\\3x}.$ Prove the inequality $x(a^{4}+b^{4})+2ya^{2}b^{2}\ge\ (x+y)(a^{3}b+ab^{3}).$ Ion Bursuc

$\iff x(a-b)^4+ab(3x-y)(a-b)^2 \ge 0$

If $ab<0?$

Generalization 2 Let $a,b,x,y$ be real numbers with ${0<(n-1)^{2}y\leq\\(2n-1)x}.$ Prove the inequality $x(a^{2n}+b^{2n})+2ya^{n}b^{n}\ge\ (x+y)(a^{2n-1}b+ab^{2n-1}).$ Ion Bursuc

Please see this: $f=\frac{1}{8}((4a+b)^2+15b^2)(a-b)^2 = 2a^4+2a^2b^2+2b^4-3a^3b-3ab^3\geq 0.$ over real numbers. and see this( use agl2010_lsos program found out): $f=(a-b)^2(a+b)^2+\frac{1}{2}(b-a)^2(a-b)^2+\frac{1}{2}b^2(a-b)^2+\frac{1}{2}a^2(a-b)^2,$ $f=(a-b)^2(a+b)^2+\frac{1}{2}a^2(a-b)^2+\frac{1}{2}b^2(a-b)^2+\frac{1}{2}(a-b)^4,$ $f=(a-b)^2(a+b)^2+\frac{1}{4}(a-b)^4+\frac{1}{2}b^2(a-b)^2+\frac{1}{2}a^2(a-b)^2+\frac{1}{4}(b-a)^2(a-b)^2,$ $f=\frac{5}{4}(a-b)^2(a+b)^2+\frac{3}{4}(b-a)^2(a-b)^2,$ $f=\frac{5}{4}(a-b)^2(a+b)^2+\frac{1}{4}(a-b)^4+\frac{1}{2}(b-a)^2(a-b)^2,$ $f=\frac{5}{4}(a-b)^2(a+b)^2+\frac{1}{4}(b-a)^2(a-b)^2+\frac{1}{2}(a-b)^4.$ BQ

for ab<0, the inequality is trivial

Refinement and generalization Let $a,b,x,y$ be real numbers with ${0<y\leq\\3x}.$ Prove the inequality $x(a^{4}+b^{4})+2ya^{2}b^{2}\geq(x+y)(a^{3}b+ab^{3})+ {\frac{3x-y}{4}}\left(a^{2}-b^{2}\right)^{2}.$ Ion Bursuc

littletush wrote: Let $a,b$ be real numbers. Prove the inequality $$2(a^4+a^2b^2+b^4)\ge 3(a^3b+ab^3).$$ We have to prove that $4(a^4+a^2b^2+b^4)\ge 6(a^3b+b^3a)$. $a^4+a^4+a^2b^2+a^2b^2\ge 4a^3b,\ (AM-GM)$ $b^4+b^4+a^2b^2+a^2b^2\ge 4ab^3,\ (AM-GM)$ $2(a^4+b^4)\ge 2(a^3b+ab^3)\ (Rearrangements)$

$a^{4}+a^{2}b^{2}\ge 2a^{3}b,\ (AM-GM)$ (1) $b^{4}+a^{2}b^{2}\ge 2b^{3}a,\ (AM-GM)$ (2) $(a^{4}+b^{4})\ge (a^{3}b+ab^{3})\ )$ (3) (1)+(2)+(3)

a^4-(a^2)(b^2)+b^4=(a^2-ab+b^2)(a^2+ab+b^2). Now, a^2+ab+b^2>=3ab and a^2-ab+b^2>=(a^2+b^2)/2 (in fact simplifying and calculating gives (a-b)^2>=0) Multiply the two equations....

LHS-RHS=$(a-b)^2(2(a^2+b^2)+ab) \geq 0$

littletush wrote: Let $a,b$ be real numbers. Prove the inequality $$2(a^4+a^2b^2+b^4)\ge 3(a^3b+ab^3).$$ $$2(a^4+a^2b^2+b^4)-3(a^3b+ab^3)=\frac{1}{4}(a-b)^2\{5(a+b)^2+3(a-b)^2\}\geq 0.$$

littletush wrote: Let $a,b$ be real numbers. Prove the inequality $$2(a^4+a^2b^2+b^4)\ge 3(a^3b+ab^3).$$ Easily proved by substituting for either $a+b, ab$ or $a+b, a-b$.

littletush wrote: Let $a,b$ be real numbers. Prove the inequality $$2(a^4+a^2b^2+b^4)\ge 3(a^3b+ab^3).$$ $$2(a^4 + a^2b^2 + b^4) - 3(a^3b + ab^3) = (a-b)^2(2a^2 + ab + 2b^2) = (a-b)^2 \left \{ 2\left (a + \frac{b}{4} \right )^2 + \frac{15}{8}b^2 \right \} \geq 0$$

OK, so we have a^4 + b^4 >= a^3b + b^3a (really trivial, just put everything on left and you will easily prove this as you will get (a-b)(a^3-b^3) = (a-b)^2*(a^2 + ab + b^2) > = 0). Also, we have a^2 + b^2 >= 2ab, (a^2 + b^2)^2 >= 2ab(a^2 + b^2) = 2a^3b + 2b^3a Combining these two inequalities we get the solution. Man, this was easy! I don't know why is this is in High School Olympiads, but is definitely far from being a challenge, even for a beginner like me

We may assume that $a\geq 0, b\geq 0$. By AM-GM, $a^4+a^2b^2\geq 2a^3b$, $b^4+a^2b^2\geq 2ab^3$ and $\frac{3}{4}a^4+\frac{1}{4}b^4\geq a^3b$, $\frac{1}{4}a^4+\frac{3}{4}b^4\geq ab^3$ Summing-up theses inequalities, we're done