Find minimize of function $P=5(a^4+b^4+a^2b^2)-3(a^2+b^2)+4$

Question

Let $a,b\in \mathbb{R}$ such that $a+b\ge 4$. Find minimize of function $$P=5(a^4+b^4+a^2b^2)-3(a^2+b^2)+4$$

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Let $a=b=2\rightarrow P=220$ so we will prove $P\ge 220$

Or $$5(a^4+b^4+a^2b^2)-3(a^2+b^2)\ge 216(1)$$

We have: $$\text{L.H.S}_{(1)}\ge5(a^4+b^4+a^2b^2)-3\cdot \frac{\left(a+b\right)^2\left(a^2+b^2\right)}{16}$$

Or we will prove $$5(a^4+b^4+a^2b^2)-3\cdot \frac{\left(a+b\right)^2\left(a^2+b^2\right)}{16}\ge \frac {27(a+b)^4}{32}$$

It is my try, i think it isnot natural and i dont know it is true or wrong.Pls checks it for me and gives me another way

Answer 1

By your work we need to prove that $$5(a^4+a^2b^2+b^4)-3(a^2+b^2)+4\geq220$$ and since $$a^4+a^2b^2+b^4\geq\frac{3}{4}(a^2+b^2)^2$$ it's $$(a^2-b^2)^2\geq0,$$ it's enough to prove that $$\frac{15}{4}(a^2+b^2)^2-3(a^2+b^2)+4\geq220$$ or $$15(a^2+b^2)^2-12(a^2+b^2)-864\geq0$$ or $$(15(a^2+b^2)+108)(a^2+b^2-8)\geq0$$ or $$a^2+b^2\geq8,$$ which is true by C-S: $$a^2+b^2=\frac{1}{2}(1+1)(a^2+b^2)\geq\frac{1}{2}(a+b)^2\geq8$$ and we are done!