Find the minimum of $\frac{a^3+2020}{b}+\frac{b^3+2020}{a}$

Question

Problem says:

Let $a,b>0$ and $2(a^2+b^2)-(a+b)=2ab$

Find the minimum of

$$\frac{a^3+2020}{b}+\frac{b^3+2020}{a}$$

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The things I have done:

$$\begin{align}\frac{2ab+(a+b)}{2}≥2ab \end{align}$$

$$\begin{align} &\implies 2ab+(a+b)≥4ab \\ &\implies a+b≥2ab \\ &\implies \frac {ab}{a+b}≤\frac 12\end{align}$$

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$$\begin{align}&2(a^2+b^2)-(a+b)=2ab, ~ab>0\end{align}$$

$$\begin{align}&\implies 2\left((a+b)^2-2ab\right)-(a+b)-2ab=0\\ &\implies 2(a+b)^2-4ab-(a+b)-2ab=0 \\ &\implies 2(a+b)^2-(a+b)-6ab=0 \\ &\implies \frac 13(a+b)-\frac{ab}{a+b}-\frac 16=0\\ &\implies \frac{ab}{a+b}=\frac 13(a+b)-\frac 16≤\frac 12\\ &\implies a+b≤2\\ &\implies 2≥a+b≥2ab \\ &\implies 0<ab≤1 \end{align}$$

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$$\begin{cases}\frac {a^3}{b}+\frac{b^3}{a} ≥2ab \\ 2020 \left(\frac 1a+\frac 1b \right)≥\frac{4040}{\sqrt{ab}}\end{cases}$$

$$\begin{align}\implies &\frac{a^3+2020}{b}+\frac{b^3+2020}{a}\\ &≥2ab+\frac{4040}{\sqrt{ab}} \end{align}$$

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Let $n:=ab$, then we have

$$f(n)=2n^2+\frac{4040}{n},~ 0<n≤1$$

$$\begin{align}f(n)-4042&=2n^2+\frac{4040}{n}-4042\\ &=\frac{2n^3-4042n+4040}{n}\\ &=\frac{2(n-1)(n^2+n-2020)}{n}\\ &≥0 \end{align}$$

$$\begin{align} &\implies f(n)-4042≥0 \\ &\implies f(n)≥4042,~ 0<n≤1\end{align}$$

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$$\begin{align}\frac{a^3+2020}{b}+\frac{b^3+2020}{a}\\ ≥2ab+\frac{4040}{\sqrt{ab}}≥4042\end{align}$$

$$\min\left\{\frac{a^3+2020}{b}+\frac{b^3+2020}{a},~{\large{\mid}} 2(a^2+b^2)-(a+b)=2ab ∧ ~a>0∧b>0\right\}=4042 ~ \text{at}~ ab=1$$

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$$\begin{cases}ab=1 \\ a+b=2 \end{cases} \implies a=b=1$$

$$\min\left\{\frac{a^3+2020}{b}+\frac{b^3+2020}{a},~{\large{\mid}} 2(a^2+b^2)-(a+b)=2ab ∧ ~a>0∧b>0\right\}=4042 ~ \text{at}~ a=b=1.$$

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• Cyclicity/symmetry argument (?)

Let,

$$f(a,b)=\frac{a^3+2020}{b}+\frac{b^3+2020}{a}$$

Then suppose that, $f(a,b)$ gets its minimum value at the point $a=m$. Substitution $a\longmapsto b$ shows that, $f(a,b)$ also gets its minimum value at the point $b=m$. This means, we have $a=b.$

I see that $$2(a^2+b^2)-(a+b)=2ab$$ is also simmetric/cyclic.

We get,

$$\begin{align}2(a^2+b^2)-(a+b)=2ab \end{align}$$

$$\begin{align}&\implies 4a^2-2a-2a^2=0 \\ &\implies a=b=1\end{align}$$

$$\begin{align}\min \left\{f(a,b) \mid 2(a^2+b^2)-(a+b)=2ab ∧ ~a>0∧b>0\right\}=4042.\end{align}$$

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Question:

• What are the points that are not rigorously correct (I refer to both proofs) in the things I do?

Please, don't post the correct solution.

Thank you for reviewing.

Answer 1

Your cyclic argument is wrong. It should be

If $f(a, b)$ attains its minimum at $(m, n)$, then it also attains its minimum at $(n, m)$

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Your first argument looks correct. No issues that I can spot thus far.

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Here's a slightly more straightforward way of showing that the condition implies $ab\leq1$ (Your approach, though correct, seems meandering/unclear what it's aiming towards.)

Let $ S = a+b$, $P = ab$. (For symmetric/cyclic equations, these are good to define.)
We have $ S^2 \geq 4P$ since $a, b > 0$.
The condition states that $2(S^2 - 2P)- S = 2P \Rightarrow 2S^2 - S = 6P$.
Hence $ 1.5 S^2 \geq 6P = 2S^2 - S \Rightarrow 2S - S^2 \geq 0 \Rightarrow 2 \geq S \geq 0 $.
Since $ S^2 \geq 4P$, thus $\Rightarrow 1 \geq P$.