Given a positive number $t$, prove that
$$7> \frac{4\,t^{3}+ 56\,t^{2}+ 191\,t+ 188}{\sqrt{t^{2}+ 2\,t+ 8}\left ( 7\sqrt{t^{2}+ 2\,t+ 8}+ 2(6\,t- 2+ t^{2}) \right )}$$
Original problem is: Given two positive numbers $x,\,y$ so that $x^{2}+ 2\,y^{2}= \frac{8}{3}$. Prove that
$$v(x)\equiv v= 7(x+ 2\,y)- 4\sqrt{x^{2}+ 2\,xy+ 8\,y^{2}}\leqq 8$$
1st solution: (with Cauchy-Schwarz) We need to prove $v(\!x\!)\leqq 7(x+ 2\,y)- (3 x+ 10\,y)= 4(x+ y)$.
Then $4(x+ y)\leqq 4\sqrt{\left ( 1+ \frac{1}{2} \right )(x^{2}+ 2\,y^{2})}= 8\,\therefore\,v(x)\leqq 8$ / q.e.d
The 2rd solution is my one without C-S :
$$x^{2}+ 2\,y^{2}= \frac{8}{3}\,\therefore\,(x- 2\,y)\left ( x- \frac{4}{3} \right )\geqq 0,\,2\,{y}'= -\,\frac{x}{y}$$
Thus, we have
$${v}'(x)= 7(1+ 2\,{y}')- \frac{4(2\,x+ 2\,y+ 2\,x{y}'+ 16\,y{y}')}{2\sqrt{x^{2}+ 2\,xy+ 8\,y^{2}}}=$$
$$= 7\left ( 1- \frac{x}{y} \right )- \frac{2\left ( 2\,x+ 2\,y- \frac{x^{2}}{y}- 8\,x \right )}{\sqrt{x^{2}+ 2\,xy+ 8\,y^{2}}}= 7\left ( 2- \frac{x}{y} \right )+ \frac{2(\,6xy- 2\,y^{2}+ x^{2})}{y\sqrt{x^{2}+ 2\,xy+ 8\,y^{2}}}- 7=$$
$$= \frac{1}{y}\left \{ -7(x- 2\,y)+ \frac{4(6\,xy- 2\,y^{2}+ x^{2})^{2}- 49\,y^{2}(x^{2}+ 2\,xy+ 8\,y^{2})^{2}}{\sqrt{x^{2}+ 2\,xy+ 8\,y^{2}}\left ( 2(6\,xy- 2\,y^{2}+ x^{2})+ 7\,y\sqrt{x^{2}+ 2\,xy+ 8\,y^{2}} \right )} \right \}=$$
$$= -\,\frac{x- 2\,y}{y}\left \{ \underbrace{7- \frac{4\,x^{3}+ 56\,x^{2}y+ 191\,xy^{2}+ 188\,y^{3}}{\sqrt{x^{2}+ 2\,xy+ 8\,y^{2}}\left ( 2(6\,xy- 2\,y^{2}+ x^{2})+ 7\,y\sqrt{x^{2}+ 2\,xy+ 8\,y^{2}} \right )}}_{> 0\,(we\,need\,to\,prove\,that\,it\,holds\,!)} \right \}$$
Indeed, for $t= -\,2\,{y}'= \frac{x}{y}> 0$, we have
$$\frac{\mathrm{d} }{\mathrm{d} t}\left ( \underbrace{2(6\,t- 2+ t^{2})+ 7\sqrt{t^{2}+ 2\,t+ 8}}_{> 0\,\because\,its\,derivative\,is\,positve\,and\,t> 0} \right )= \frac{14(t+ 1)}{2\sqrt{t^{2}+ 2\,t+ 8}}+ 4(t+ 3)> 0$$
Or
$$14(6 t- 2+ t^{2})\sqrt{t^{2}+ 2 t+ 8}> 4 t^{3}+ 56 t^{2}+ 191 t+ 188- 49(t^{2}+ 2 t+ 8)\,(\!original.problem\!)$$
My solution in VMF : (and I'm looking forward to seeing a nicer one(s), thanks for your interests !)
For $0< t\leqq 2$
$$14(6\,t- 2+ t^{2})\sqrt{t^{2}+ 2\,t+ 8}\geqq 4\,t^{3}+ 241\,t^{2}+ 192\,t+ 188- 49(t^{2}+ 2\,t+ 8)>$$
$$> 4\,t^{3}+ 56\,t^{2}+ 191\,t+ 188- 49(t^{2}+ 2\,t+ 8)$$
For $t\geqq 2$
$$14(6\,t- 2+ t^{2})\sqrt{t^{2}+ 2\,t+ 8}\geqq 4\,t^{3}+ 56\,t^{2}+ 562\,t+ 188- 49(t^{2}+ 2\,t+ 8)>$$
$$> 4\,t^{3}+ 56\,t^{2}+ 191\,t+ 188- 49(t^{2}+ 2\,t+ 8)$$
Therefore $(x- 2\,y){v}'(x)\leqq 0\,\therefore\,\left ( x- \frac{4}{3} \right ){v}'(x)\leqq 0\,\therefore\,v(x)\leqq v\left ( \frac{4}{3} \right )= 8$ / q.e.d
Best Answer
Since by C-S $$\sqrt{t^2+2t+8}=\frac{\sqrt{(9+7)((t+1)^2+7)}}{4}\geq\frac{3(t+1)+7}{4}=\frac{3t+10}{4},$$ it's enough to prove that $$\frac{7(3t+10)}{4}\left(\frac{7(3t+10)}{4}+2(t^2+6t-2)\right)>4t^3+56t^2+191t+188$$ or $$104t^3+1113t^2+2908t+772>0,$$ which is obvious.