Given two positive numbers $x,\,y$ so that $32\,x^{6}+ 4\,y^{3}= 1$. Prove that
$$p(x)\equiv p= \frac{(2\,x^{2}+ y+ 3)^{5}}{3(x^{2}+ y^{2})- 3(x+ y)+ 2}\leqq 2048$$
My solution in VMF : (and I'm looking forward to seeing a nicer one(s), thanks for your interests !)
$$32\,x^{6}+ 4\,y^{3}= 1\,\therefore\,(y- 2\,x^{2})\left ( x- \frac{1}{2} \right )\leqq 0,\,\left ( x- \frac{1}{2} \right )\left ( y- \frac{1}{2} \right )\leqq 0,\,{y}'= -\,\frac{16\,x^{5}}{y^{2}}$$
Thus, we have
$${p}'(x)=$$
$$= \frac{{\left (\!(\!2 x^{2}+ y+ 3\!)^{5}\!\right )}'\{\!3(\!x^{2}+ y^{2}\!)- 3(x+ y)+ 2\!\}- {\left (\!3(\!x^{2}+ y^{2}\!)- 3(x+ y)+ 2\!\right )}'(\!2 x^{2}+ y+ 3\!)^{5}}{\left (\!3(\!x^{2}+ y^{2}\!)- 3(x+ y)+ 2\!\right )^{2}}=$$
$$\frac{5(\!2 x^{2}+ y+ 3\!)^{4}\left (\!4 x- \frac{16 y^{5}}{x^{2}}\!\right )\{\!3(\!x^{2}+ y^{2}\!)- 3(x+ y)+ 2\!\}- \left (\!6 x- 3- \frac{96 x^{5}}{y}+ \frac{48 x^{5}}{y^{2}}\!\right )(\!2 x^{2}+ y+ 3\!)^{5}}{\left (\!3(\!x^{2}+ y^{2}\!)- 3(x+ y)+ 2\!\right )^{2}}$$
$$= \frac{\left ( 20\,x(y- 2\,x^{2})(y+ 2\,x^{2})\{\!3(x^{2}+ y^{2})- 3(x+ y)+ 2\!\}- 3[x^{5}(16- 32\,y)+ y^{2}(2\,x- 1)] \right )}{y^{2}\left ( 3(x^{2}+ y^{2})- 3(x+ y)+ 2 \right )^{2}}$$
$$\times (2\,x^{2}+ y+ 3)^{4}$$
Using derivatives, the equation of the tangent line can be stated as follows:
$$p(x)- p\left ( \frac{1}{2} \right )= {p}'(x)\left ( x- \frac{1}{2} \right )\leqq 0\,(\!easy\,to\,see\,immediately\,!\!)\,\therefore\,p(x)\leqq p\left ( \frac{1}{2} \right )\leqq 2048$$
Given two positive numbers $x,\,y$ so that $32\,x^{6}+ 4\,y^{3}= 1$. Prove that $\frac{(2\,x^{2}+ y+ 3)^{5}}{3(x^{2}+ y^{2})- 3(x+ y)+ 2}\leqq 2048$ .
a.m.-g.m.-inequalityderivativesinequalitysubstitutiontangent line
Related Question
- Given a positive number $t$, prove that $7>\frac{4t^{3}+56t^{2}+191t+188}{\sqrt{t^{2}+2t+8}\left (7\sqrt{t^{2}+2t+8}+2(6t-2+t^{2})\right )}$ .
- Given three positive numbers $a,b,c$ so that $abc= 1$. Prove $(a-1+\frac{1}{b})(b-1+\frac{1}{c})(c-1+\frac{1}{a})\leqq\frac{2}{a+b+c-1}$ .
- Given three non-negative numbers $x,y,z$ so that $x+y+z=100$ and $x,y,z\notin (\!33,34\!)$. Prove that $xyz\leqq 33\cdot 34\cdot 33$ .
Best Answer
By AM-GM $$1=32x^6+4y^3=32x^6+2\cdot\frac{1}{2}+4y^3+2\cdot\frac{1}{2}-2\geq$$ $$\geq3\sqrt[3]{32x^6\cdot\left(\frac{1}{2}\right)^2}+3\sqrt[3]{4y^3\cdot\left(\frac{1}{2}\right)^2}-2=6x^2+3y-2,$$ which gives $$2x^2+y\leq1.$$ Id est, $$\frac{(2x^2+y+3)^5}{3(x^2+y^2)-3(x+y)+2}\leq\frac{4^5}{3(x^2+y^2)-3(x+y)+\frac{3}{2}+\frac{1}{2}}=$$ $$=\frac{1024}{3\left(x^2+y^2-x-y+\frac{1}{2}\right)+\frac{1}{2}}=\frac{1024}{3\left(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2\right)+\frac{1}{2}}\leq\frac{1024}{\frac{1}{2}}=2048$$ and we are done!