Note that we can write $a$ as $a=18-\left(bc+cd+db+\frac{1}{ab^2c^2d^2}\right)\leq18-\left(\frac{4}{\sqrt[4]{a}}\right)$ since we need to find $a_{\text{max}}$ , minimizing this part $\left(bc+cd+db+\frac{1}{ab^2c^2d^2}\right)$ will work, hence the above part which is true by AM-GM, we need to solve this $$a+4a^{-\frac{1}{4}}-18=0$$which yields $a=16$ as the value.

Using AM-GM on the last four terms, we obtain \[ 18\ge a+ 4a^{-\frac14}\iff t^4 +\frac4t-18\le 0 \qquad\text{where}\qquad a=t^4. \]From here, $t\le 2$ and therefore $a\le 16$. We now give an example attaining the equality. Note that if $b=c=d$ and $b^2 = (ab^6)^{-1}\iff b^8 = \frac1a$, we enjoy equality. Namely, $a=16$ is indeed the maximal value, concluding the case.

If the positive real numbers $a,b$ satisfy the equality $2a+b+\frac{4}{ab}=10. $Find the maximum value of $a$. If the positive real numbers $a,b,c,d$ satisfy the equality $a+bc+bd+cd+\frac{1}{ab^2c^2d^2}=18.$ Find the maximum value of $a$. Funny twins.

Let $a,b>0$ and $2a+b+\frac{4}{ab }=6. $Prove that$$2a^2+b^2 \leq 6$$$$2a^2+\frac{1}{b^2}\leq \frac{9}{4}$$

Let $a,b\in (0,1]$ and $2a+b+\frac{4}{ab}=10. $ Prove that$$a+2b\leq \frac{5}{2}$$$$a+ \frac{1}{b}\leq 2+\frac{\sqrt 3}{2}$$

sqing wrote: If the positive real numbers $a,b$ satisfy the equality $2a+b+\frac{4}{ab}=10. $Find the maximum value of $a$. $$10=2a+b+\frac{4}{ab}\geq 10\sqrt[10]{\left(\frac{2a}{8}\right)^8\cdot b\cdot \frac{4}{ab}}=10\sqrt[10]{\left(\frac{a}{4}\right)^7}$$$$a\leq 4$$Equality when $a=4,b=1.$ sqing wrote: Let $a,b\in (0,1]$ and $2a+b+\frac{4}{ab}=10. $ Prove that$$a+2b\leq \frac{5}{2}$$Equality when $a=\frac{1}{2},b=1.$ $$a+ \frac{1}{b}\leq 2+\frac{\sqrt 3}{2}$$Equality when $a=1,b=4-2\sqrt 3 .$

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