Any Classical Inequalities Approach?

Relevant wiki: Trigonometric R method

Thus , $a^2+2b^2=2$ and then rewrite it we got :

$\frac{a^2}{2}+b^2=1$

Let $\frac{a}{\sqrt{2}}=\sin x ; b=\cos x$

So $P=\sqrt{2}\sin x+3\cos x-2$

By R method , the equation has the solution when : $(P+2)^2 \leq \sqrt{(\sqrt2)^2 + 3^2} = \sqrt{11}$

$-2-\sqrt{11}\leq P\leq -2+\sqrt{11}$

So the sum is $(-2-\sqrt{11}) + (-2+\sqrt{11}) = \boxed{-4}$ .

If $(a,b)$ satisfy $a^2 + 2b^2 = 2$ then so do $(-a,-b)$ as $(-a)^2 + 2(-b)^2 = a^2 + 2b^2 = 2$ . Let $(a,b)$ be the values which makes $P$ at its maximum. Let $(x,y)$ be values that satisfy $a^2 + 2b^2 = 2$ and let $P' = x+3y - 2$ . Since $(a,b)$ is gives the maximum value of $P$ , we must have $\begin{aligned} P &\geq P' \\ a + 3b &\geq x+3y \\-x + 3(-y) &\geq -a + 3(-b) \\ -x + 3(-y) - 2 &\geq -a + 3(-b) - 2 \end{aligned}$ . This means that for any $(-x,-y)$ , $-x +3(-y) - 2$ will always be greater than $-a + 3(-b) - 2$ (with equality holding when $(x,y) = (a,b)$ ). Thus when the $(a,b)$ gives the maximum value of $P$ , $(-a,-b)$ will give the minimum value. The sum of $P$ for these two values are $a + 3b - 2 + (-a) + 3(-b) - 2 = \boxed{-4}$

Relevant wiki: Cauchy-Schwarz Inequality

One more solution wouldn't hurt. By the cauchy-schwarz inequality,

$(1^2) +(\frac{3}{\sqrt{2}})^2)((a)^2+(\sqrt{2}b)^2) \geq (a+(\sqrt{2})(\frac{3}{\sqrt{2}})(b))^2$ , giving us

$11 \geq (a+3b)^2$

This implies

$\sqrt{11} \geq a+3b \geq -\sqrt{11}$ .

Subtracting $2$ from both sides gives us

$\sqrt{11} -2 \geq a+3b-2 \geq -\sqrt{11}-2$ .

Therefore the maximum is $\sqrt{11}-2$ , minimum is $-\sqrt{11}-2$ . Our answer is then $\sqrt{11}-2-\sqrt{11}-2 = \boxed{-4}$ .

Shortcut based on symmetry

If we replace $(a,b)$ by $(-a,-b)$ , the value of $a^2 + 2b^2 = 2$ remains the same, while the value of $P$ is replaced by $-4-P$ . If $(a^\star,b^\star)$ gives a maximum then $(-a^\star,-b^\star)$ gives the minimum, and $P_{min} = -4-P_{max}$ . Thus $P_{min} + P_{max} = \boxed{-4}.$ without knowing any further details!

Relevant wiki: Trigonometry R method

Similar solution as @MS HT's, different presentation.

$\begin{aligned} a^2 + 2b^2 & = 2 & \small \color{#3D99F6}{\text{Divide both sides by }2} \\ \frac {a^2}2 + b^2 & = 1 & \small \color{#3D99F6}{\text{Using the identity }\sin^2 \theta + \cos^2 \theta = 1} \end{aligned}$

$\implies \begin{cases} \dfrac a{\sqrt 2} = \sin \theta & \implies a = \sqrt 2 \sin \theta \\ b = \cos \theta \end{cases}$

Then, we have:

$\begin{aligned} P & = a+3b-2 \\ & = \sqrt 2 \sin \theta + 3 \cos \theta - 2 \\ & = \sqrt{11}\left(\sqrt {\frac {2}{11}}\sin \theta + \frac 3{\sqrt{11}} \cos \theta \right) - 2 \\ & = \sqrt{11} \color{#3D99F6}{\sin \left(\theta + \tan^{-1} \frac 3{\sqrt 2} \right)} - 2 \end{aligned}$

We note that $P$ is maximum and minimum when $\color{#3D99F6}{\sin \left(\theta + \tan^{-1} \frac 3{\sqrt 2} \right)}$ is maximum $=1$ and minimum $=-1$ respectively. Therefore, $P_{max} = \sqrt{11} - 2$ and $P_{min} = -\sqrt{11} - 2$ , and $P_{max} + P_{min} = \boxed{-4}$ .

Natural approach to this problem would be trigonometric substitution , but just for sake of variety, here's another method.

We have to find minimum and maximum value of $P$

From given constraint, $a^2 = 2(1-b^2)$

Also, $(P -3b+2)^2 = a^2$

Therefore, $(P-3b+2)^2 =2(1-b^2)$ .

Rearranging a bit, we get $11b^2 -6b(P+2) +P^2 +4P +2 =0$

Let us consider the above equation as a quadratic in $b$ .

Since $b \in R$ , the discriminant of the quadratic equation $\ge0$

$\implies (-6(P+2))^2 -4(11)(P^2 + 4P +2) \ge 0$

$\boxed{\implies -2-\sqrt{11}\leq P\leq -2+\sqrt{11}}$

Therefore final answer is $-4$ .

Put a= cos and b =sin