Find maximize of $P=\frac{x\sqrt{yz}}{\sqrt{x^2+1}\sqrt[4]{\left(y^2+4\right)\left(z^2+9\right)}}$

Question

Let $x,y,z\in \mathbb{R^+}$ such that $6x+3y+2z=xyz$. Find maximize of $$P=\frac{x\sqrt{yz}}{\sqrt{x^2+1}\sqrt[4]{\left(y^2+4\right)\left(z^2+9\right)}}$$

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We will prove $$P\le \sqrt{\frac{16}{27}}$$

$$\Leftrightarrow \frac{1}{\sqrt{1+\frac{1}{x^2}}\sqrt[4]{\left(1+\frac{4}{y^2}\right)\left(1+\frac{9}{z^2}\right)}}\le \sqrt{\frac{16}{27}}$$

$$\Leftrightarrow \frac{1}{\sqrt[4]{\frac{\left(2x+y\right)^3\left(3y+2z\right)^2\left(3x+z\right)^3}{x^6y^5z^5}}}\le \sqrt{\frac{16}{27}}$$

$$\Leftrightarrow \left(2x+y\right)^3\left(3y+2z\right)^2\left(3x+z\right)^3\left(xyz\right)^4\ge \frac{729}{256}x^6y^5z^5\left(6x+3y+2z\right)^4$$

I don't know how to solve it now. Help me.

Answer 1

Also, $uvw$ helps.

Indeed, in my previous solution it's enough to prove that: $$64(2a+b)^3(2a+c)^3(b+c)^2\geq729(2a+b+c)^4a^2bc.$$ Now, let $b+c=2u$ and $bc=v^2$, where $v>0$.

Thus, by AM-GM $u\geq v$ and we need to prove that $f(u)\geq0,$ where $$f(u)=16u^2(4a^2+4ua+v^2)^3-729(a+u)^4a^2v^2.$$ But, $$f'(u)=32u(4a^2+4ua+v^2)^3+192u^2a(4a^2+4ua+v^2)^2-4\cdot729(a+u)^3a^2v^2=$$ $$=4(512ua^6+9(256u^2-81v^2)a^5+3u(1024u^2-601v^2)a^4+5u^2(256u^2-207v^2)a^3+$$ $$+3uv^2(13u^2+32v^2)a^2+144u^2v^4a+8uv^6)>0,$$ which says that $f$ increases.

Thus, it's enough to prove our inequality for a minimal value of $u$, which happens for $u=v$,

id est, for $b=c$.

Since our inequality is homogeneous, it's enough to assume that $b=c=1$ and we need to prove that: $$16(4a^2+4a+1)^3\geq729(a+1)^4a^2$$ or $$4(2a+1)^3\geq27(a+1)^2a,$$ which is true by AM-GM: $$4(2a+1)^3=4\left(a+2\cdot\frac{a+1}{2}\right)^3\geq4\left(3\sqrt[3]{a\left(\frac{a+1}{2}\right)^2}\right)^3=27(a+1)^2a.$$

Answer 2

Now, let $x=2\sqrt2a$, $y=2\sqrt2b$ and $z=3\sqrt2c$.

Thus, we need to prove that: $$64(2a+b)^3(2a+c)^3(b+c)^2\geq729(2a+b+c)^4a^2bc.$$ Also, let $a=t\sqrt{bc}$.

Thus, $$64(2a+b)^3(2a+c)^3(b+c)^2-729(2a+b+c)^4a^2bc=$$ $$=16(256b^2-217bc+256c^2)a^6+96(64b^3-51b^2c-51bc^2+64c^3)a^5+$$ $$+24(b+c)^2(128b^2-354bc+128c^2)a^4+8(b+c)^3(64b^2-217bc+64c^2)a^3+$$ $$+3bc(b+c)^2(13b^2+282bc+13c^2)a^2+384b^2c^2(b+c)^3a+64b^3c^3(b+c)^2\geq$$ $$\geq16\cdot295bca^6+96\cdot26\sqrt{b^3c^3}a^5-8544b^2c^2a^4-5696\sqrt{b^5c^5}a^3+$$ $$+12\cdot308b^3c^3a^2+384\cdot8\sqrt{b^7c^7}a+256b^4c^4=$$ $$=16b^4c^4(295t^6+156t^5-534t^4-356t^3+231t^2+192t+16)=$$ $$=16b^4c^4(t-1)^2(5t+4)(59t^3+102t^2+51t+4)\geq0.$$