knm2608 wrote: If the positive reals $x,y,z$ satisfy $x^2+y^2+z^2=x+y+z$. Prove that $$\displaystyle\frac{x+1}{\sqrt{x^5+x+1}}+\frac{y+1}{\sqrt{y^5+y+1}}+\frac{z+1}{\sqrt{z^5+z+1}}\geq 3.$$ JBMO Shortlist 2016, A4

sqing wrote: knm2608 wrote: If the positive reals $x,y,z$ satisfy $x^2+y^2+z^2=x+y+z$. Prove that $$\displaystyle\frac{x+1}{\sqrt{x^5+x+1}}+\frac{y+1}{\sqrt{y^5+y+1}}+\frac{z+1}{\sqrt{z^5+z+1}}\geq 3.$$ JBMO Shortlist 2016, A4 Yes. Thank you.

This problem was proposed by me. And also it should be 'non-negative real numbers' not 'positive real numbers'.

knm2608 wrote: If the non-negative reals $x,y,z$ satisfy $x^2+y^2+z^2=x+y+z$. Prove that $$\displaystyle\frac{x+1}{\sqrt{x^5+x+1}}+\frac{y+1}{\sqrt{y^5+y+1}}+\frac{z+1}{\sqrt{z^5+z+1}}\geq 3.$$When does the equality occur? Proposed by Dorlir Ahmeti, Albania $$\frac{x+1}{\sqrt{x^5+x+1}} \geqslant 1 + \frac{x(1-x)}{2}$$

Hmmm.......nice one . $p^5+p+1=(p^2+p+1)(p^3–p^2+1)$ ,$ p>1$ by roots of unity $w^2+w+1=w^5+w^2+1$ And as follows above @hmmmm

JBMO SL 2016 A4 wrote: If the non-negative reals $x,y,z$ satisfy $x^2+y^2+z^2=x+y+z$. Prove that $$\displaystyle\frac{x+1}{\sqrt{x^5+x+1}}+\frac{y+1}{\sqrt{y^5+y+1}}+\frac{z+1}{\sqrt{z^5+z+1}}\geq 3.$$When does the equality occur? Solution. $$x^5+x+1=x^5-x^2+x^2+x+1=x^2(x^3-1)+x^2+x+1=(x^2+x+1)(x^2(x-1)+1)=(x^2 + x + 1)(x^3- x^2 + 1)$$Using the AM-GM inequality we have that: $$\frac{x+1}{\sqrt{(x^5+x+1)}}\geq \frac{2(x+1)}{x^3+x+2}=\frac{2}{x^2-x+2}$$Now it is enough to show that $\frac{2}{x^2-x+2}+\frac{2}{y^2-y+2}+\frac{2}{z^2-z+2}\geq 3\iff \frac{1}{x^2-x+2}+\frac{1}{y^2-y+2}+\frac{1}{z^2-z+2} \geq \frac{3}{2}$ But, by using the CS-Engel inequality we have that: $$\frac{1}{x^2-x+2}+\frac{1}{y^2-y+2}+\frac{1}{z^2-z+2} \geq \frac{9}{6} = \frac{3}{2}.\blacksquare$$

@above beautiful solution! Congratulations!

Magnificent. Notice that $x^5+x+1=(x^2+x+1)(x^3-x^2+1)$, so by AM-GM $$(x+1)(x^2-x+2)=x^3+x+2=(x^3-x^2+1)+(x^2+x+1)\geq2\sqrt{(x^2+x+1)(x^3-x^2+1)}=2\sqrt{x^5+x+1}$$. Hence $$\sum_{cyc}\frac{x+1}{\sqrt{x^5+x+1}}\geq \sum_{cyc}\frac{2(x+1)}{(x+1)(x^2-x+2)}=\sum_{cyc}\frac{2}{x^2-x+2} \stackrel{Titu}{\geq} \frac{18}{x^2+y^2+z^2-x-y-z+6}=3$$As desired. For an equality we should have an equality when using AM-GM, i.e. $$x^2+x+1=x^3-x^2+1$$$$x(x^2-2x-1)=0$$The non-negative solutions to these are $x=0$ and $x=1+\sqrt{2}$. So each of $x,y,z$ is either $0$ or $1+\sqrt{2}$. When using Titu for an equality we should have $$x^2-x+2=y^2-y+2$$$$(x-y)(x+y-1)=0$$Since for $x,y$ being either $0$ or $1+\sqrt{2}$ their sum cannot be $1$, $x=y$. Similarly we get $y=z$, so $x=y=z$. Substitute into the given condition $$3x^2=3x$$$$x(x-1)=0$$So $x,y,z$ are either all $1$ or all $0$. $1$ obviously does not yield an equality, meanwhile $0$ surely does. So the only equality case is $x=y=z=0$. $\blacksquare$

knm2608 wrote: If the non-negative reals $x,y,z$ satisfy $x^2+y^2+z^2=x+y+z$. Prove that $$\displaystyle\frac{x+1}{\sqrt{x^5+x+1}}+\frac{y+1}{\sqrt{y^5+y+1}}+\frac{z+1}{\sqrt{z^5+z+1}}\geq 3.$$When does the equality occur? Proposed by Dorlir Ahmeti, Albania Note that $$\frac{1}{\sqrt{x^5+x+1}} \ge 1-\frac{x}{2}$$for all positive reals(It is the tangent line to $\frac{1}{\sqrt{x^5+x+1}}$ at $x=0$). Hence we can conclude that $$\frac{x+1}{\sqrt{x^5+x+1}}+\frac{y+1}{\sqrt{y^5+y+1}}+\frac{z+1}{\sqrt{z^5+z+1}} \ge \sum_{cyc} (x+1)\left(1-\frac{x}{2}\right)$$$$=\sum_{cyc} 1+\frac{1}{2}\sum_{cyc} x-\frac{1}{2}\sum_{cyc} x^2=3$$as desired.

Note that $x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$. So by AM-GM we have $(x^2 + x + 1) + (x^3 - x^2 + 1) \geq 2\sqrt{x^5 + x + 1} \implies \sqrt{x^5 + x + 1} \leq \frac{x^3 + x + 2}{2}$. So inequality rewrites as $\sum{\frac{x+1}{x^3 + x + 2}} \geq \frac{3}{2}$. Note that $x^3 + x + 2 = (x+1)(x^2 - x + 2)$. So it suffices to show $\sum{\frac{1}{x^2 - x + 2}} \geq \frac{3}{2}$. But that is easily proven by Cauchy $\sum{\frac{1}{x^2 - x + 2}} \geq \frac{9}{(x^2 + y^2 + z^2) - (x + y + z) + 6} = \frac{3}{2}$. Equality is satisfied when $x=y=z=0$.

Note that by AM-GM $$\sum \frac{x+1}{\sqrt{x^5+x+1}} = \sum \frac{x+1}{\sqrt{(x^2+x+1)(x^3-x^2+1)}} \geq \sum \frac 2{x^2-x+2}.$$But we obviously have $$\sum \frac 2{x^2-x+2} \geq \frac 96 = \frac 32$$by C-S and the given condition.

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Factorizing and using AM-GM, the LHS can be expressed as \[\sum_{\text{cyc}} \frac{x+1}{\sqrt{(x^2+x+1)(x^3-x^2+1)}} \ge \sum_{\text{cyc}} \frac{2(x+1)}{x^3+x+2} = \sum_{\text{cyc}} \frac{2}{x^2-x+2}.\] We can finish by using Titu's: \[\sum_{\text{cyc}} \frac{2}{x^2-x+2} \ge \frac{(\sqrt 2 + \sqrt 2 + \sqrt 2)^2}{\sum_{\text{cyc}} (x^2-x+2)} = \frac{18}{6} = 3.~\blacksquare\]

MATH1945 wrote:

(2 + 2 + 2)^2 = 36 and 36 / 6 = 6. LHS >= 6.

Lol.just note these::: $x^5+x+1=(x^3-x^2+1)(x^2+x+1)$ So by $AM-GM$ we get::: $x^3+x+2=(x+1)(x^2-x+2)\ge x^5+x+1$ Applying to all and using T2 gives the answer