Problem 268

Proof:

(1) Given $a=c$ and $a^{2}+2b^{2}+3c^{2}=4$, we substitute $c$ with $a$ to get:

$$4a^{2}+2b^{2}=4$$

This can be simplified to:

$$2a^{2}+b^{2}=2$$

By applying the AM-GM inequality, we have:

$$2 = 2a^{2} + b^{2} \geq 2\sqrt{2a^{2}b^{2}} = 2\sqrt{2}ab$$

Dividing both sides by $2\sqrt{2}$, we get:

$$ab \leq \frac{\sqrt{2}}{2}$$

Thus, we have proven that if $a=c$, then $ab\leqslant \frac{{\sqrt{2}}}{2}$. The equality holds when $b=\sqrt{2}a$, which encapsulates our final answer for part (1) as:

$$\boxed{ab \leq \frac{\sqrt{2}}{2}}$$

(2) Given $a^{2}+2b^{2}+3c^{2}=4$ and using the Cauchy-Schwarz inequality, we consider the sum of squares on the left and a sum of squares of coefficients on the right:

$$(a^{2}+2b^{2}+3c^{2})(1^{2}+\sqrt{2}^{2}+\sqrt{3}^{2}) \geqslant (a+2b+3c)^{2}$$

Simplifying the terms, we get:

$$4 \times 6 \geqslant (a+2b+3c)^{2}$$

Taking the square root of both sides, we find:

$$a+2b+3c \leqslant 2\sqrt{6}$$

This proves that $a+2b+3c\leqslant 2\sqrt{6}$, with equality holding when $a=b=c=\frac{\sqrt{6}}{3}$. Thus, the final encapsulated answer for part (2) is:

$$\boxed{a+2b+3c \leqslant 2\sqrt{6}}$$