<s>[INST] Why the maximum value of $|AB|+\sqrt{2}|CD|$ is $10$? 
Assume a Convex quadrilateral $ABCD$ a,such $|AC|=1,|BC|=3\sqrt{2}$,and $|AD|=|BD|=\frac{\sqrt{2}}{2}|AB|$, find the maximum of the value  $|AB|+\sqrt{2}|CD|$

The Book  only have answer: $10$.
I wanted to solve this by using law of cosines in triangles $ABC$and $BDC$ to determine $AB$ and $CD$,
Assume that $\angle ACB=x$,then $AB=\sqrt{1+18-6\sqrt{2}\cos{x}}=\sqrt{19-6\sqrt{2}\cos{x}}$，But I can't determine $CD$ .

 [/INST] Let $\angle ADC=x$, then $\angle BDC=90^\circ-x$. Apply the law of cosine to $\triangle ADC$ we have
$$2AD\cdot CD\cos x=AD^2+CD^2-1\tag{1}$$
and to $\triangle BCD$ we have
$$2BD\cdot CD \cos (90^\circ-x) = BD^2+CD^2-18\tag{2}.$$
Square both (1) and (2) and add them up we get
$$4AD^2\cdot CD^2 = (AD^2+CD^2-1)^2+(AD^2+CD^2-18)^2.$$
so
$$0=2AD^4+2CD^4-26(AD^2+CD^2)+145\ge (AD^2+CD^2)^2-38(AD^2+CD^2)+(1+18^2).$$
It follows that
$$AD^2+CD^2\le 19+\sqrt{19^2-1-18^2}=19+6=25.$$
Thus
$$(AD+CD)^2\le 2(AD^2+CD^2)=50$$
or equivalently
$$\frac{AB+\sqrt{2}CD}{\sqrt2}\le \sqrt{2(AD^2+CD^2)}\le \sqrt{50}.$$
So 
$$AB+\sqrt{2}CD\le 10.$$
The equality happens when $AD=CD=BD$, equivalently $\angle ACB=135^\circ$.
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