Let $\sqrt{a^2+c^2+e^2}:=S'$ and $\sqrt{b^2+d^2+f^2}:=T'$. Furthermore, define $x:=b-a$, $y:=d-c$, $z:=f-e$ and $v:=c-b$, $w:=e-d$. Then $2(ac+ce+ea)=(a+c+e)^2-(a^2+c^2+e^2)=S^2-S'^2$; likewise, $2(bd+df+fb)=T^2-T'^2$. Squaring the initial inequality and then multiplying by $2$ yields \[8S^2T^2>3(S+T)(S(T^2-T'^2)+T(S^2-S'^2))\] \[\iff 3S'^2(S+T)T+3T'^2(S+T)S>3(S+T)^2ST-8S^2T^2\] \[\iff (S+T)\left(T(3S'^2-S^2)+S(3T'^2-T^2)\right)\\>3(S+T)^2ST-8S^2T^2-(S+T)^2ST=2ST(S-T)^2.\quad (1)\] We have \[3S'^2-S^2=3(a^2+c^2+e^2)-(a+c+e)^2= (a-c)^2+(c-e)^2+(e-a)^2\\\quad \quad =(x+v)^2+(y+w)^2+(x+y+v+w)^2.\] Thus, $3S'^2-S^2>x^2+y^2+(x+y)^2=2(x^2+y^2+xy)$. A similar estimation can be obtained for $3T'^2-T^2$. Taking into account these two estimations and $T-S=x+y+z$, (1) can be sharpened as follows: \[\frac{S+T}{S}(x^2+y^2+xy)+\frac{S+T}{T}(y^2+z^2+yz)>(x+y+z)^2\] \[\iff T^2(x^2+y^2+xy)+S^2(y^2+z^2+yz)>ST(xy+yz+2xz-y^2). \quad (2)\] Now, \[T^2x^2+S^2z^2-2STxz=(xT-zS)^2, \\ T^2xy+S^2yz-ST(xy+yz)=y(T-S)(xT-zS).\] Hence (2) is equivalent to \[ (xT-zS)^2+y(T-S)(xT-zS)+y^2(T^2+S^2+TS)>0 \\ \iff \left( (xT-zS)+y\frac{T-S}{2}\right)^2+y^2\left(T^2+S^2+TS-\frac{(T-S)^2}{4}\right)>0,\] which is clearly true.

Let $a+c+e=u$, $ac+ce+ea=w$, $b+d+f=y$, $bd+df+fb=z$, $u'=u^2-3w>0$, $y'=y^2-3z>0$, and let $g(x)=3x^2-2xy+z$, $h(x)=3x^2-2xu+w$. Let $p=\frac{1}{3}(y+ \sqrt{y'})$, $r=\frac{1}{3}(y- \sqrt{y'})$ be the roots of $g(x)$, and $q=\frac{1}{3}(u+ \sqrt{u'})$, $s=\frac{1}{3}(u- \sqrt{u'})$ be the roots of $h(x)$. Since $y^2>y'$, $u^2>u'$, we have $p$, $q$, $r$, $s$ $>0$. Now we claim $(p-q)(r-s)>0$ $\Longleftrightarrow (\sqrt{y'} + \sqrt{u'} + y-u)(\sqrt{y'} + \sqrt{u'} + u-y)>0$ $\Longleftrightarrow (\sqrt{y'} + \sqrt{u'})^2-(y-u)^2>0$ $\Longleftrightarrow y^2-3z+u^2-3x+2\sqrt{u'y'} -y^2-u^2+2uy>0$ $\Longleftrightarrow 2uy+\sqrt{4u'y'}-3w-3z>0$ From Cauchy-Schwarz $\sqrt{4u'y'} = \sqrt{((f-d)^2+(f-b)^2+(d-b)^2)((c-a)^2+(e-a)^2+(e-c)^2)}$ $\ge (f-d)(c-a)+(f-b)(e-a)+(d-b)(e-c) =$ $ f(e+c-2a)+d(a+e-2c)+b(a+c-2e)$. Also, $uy=f(a+c+e)+d(a+c+e)+b(a+c+e)$, so it suffices to show $3f(e+c)+3d(a+e)+3b(a+c) -3(ac+ce+ea+bd+df+fb) >0$ $\Longleftrightarrow (f-a)(e-b)+(f-a)(c-d)+(e-b)(d-c)>0$ $\Longleftrightarrow (e-b)(f-a+d-c)>(d-c)(f-a)$, which is true since $e-b>d-c$, $f-a+d-c>f-a$. The given inequality is equivalent to $4u^2y^2>3(u+y)(uz+yw)$. From Vieta's relations on $g(x)$, $h(x)$, we have $pr=\frac{z}{3}$, $p+r=\frac{2y}{3}$, $qs=\frac{w}{3}$, $q+s=\frac{2u}{3}$, so it suffices to show $(p+r)^2(q+s)^2>(p+q+r+s)(pqs+rqs+prq+prs)$. But we have $(p+r)^2(q+s)^2-(p+q+r+s)(pqs+rqs+prq+prs) =$ $ (p-q)(r-s)(p+s)(q+r) + (pq-rs)^2>0$.

Squaring both sides, the inequality becomes \[4S^2T^2>3(S+T)(S(bd+bf+df)+T(ac+ae+ce))\] \[\Leftrightarrow ST(S+T)^2-4S^2T^2<ST(S+T)^2-3(S+T)(S(bd+bf+df)+T(ac+ae+ce))\] \[\Leftrightarrow ST(S-T)^2<(S+T)(S^2T+ST^2-3S(bd+bf+df)-T(ac+ae+ce))\] \[\Leftrightarrow ST(S-T)^2<(S+T)(T(S^2-3(ac+ae+ce))+S(T^2-3(bd+bf+df)))\] By Cauchy-Schwarz, \[RHS\ge\left(\sqrt{ST(S^2-3(ac+ae+ce))}+\sqrt{ST(T^2-3(bd+bf+df))}\right)^2=ST\left(\sqrt{\frac{(a-c)^2+(c-e)^2+(e-a)^2}2}+\sqrt{\frac{(b-d)^2+(d-f)^2+(f-b)^2}2}\right)^2.\] So it suffices to prove that \[\left(\sqrt{\frac{(a-c)^2+(c-e)^2+(e-a)^2}2}+\sqrt{\frac{(b-d)^2+(d-f)^2+(f-b)^2}2}\right)^2>(S-T)^2\] Let $b=a+k$, $c=b+l$, $d=c+m$, $e=d+n$, $f=e+p$. The inequality becomes \[(\sqrt{(k+l)^2+(m+n)^2+(k+l+m+n)^2}+\sqrt{(l+m)^2+(n+p)^2+(l+m+n+p)^2})^2>2(k+m+p)^2.\] It suffices to prove that \[\sqrt{k^2+m^2+(k+m)^2}+\sqrt{m^2+p^2+(m+p)^2}>\sqrt2(k+m+p)\] which is true because $\sqrt{k^2+m^2+(k+m)^2}=\sqrt{2(k^2+km+m^2)}>\sqrt2\left(k+\frac{m}2\right)$ and $\sqrt{m^2+p^2+(m+p)^2}=\sqrt{2(m^2+mp+p^2)}>\sqrt2\left(p+\frac{m}2\right)$.

This problem was proposed by me. This problem was so nasty that I don't like it while it is chosen for the shortlist. My original solution uses an appropriate polynomial.

Sung-yoon Kim wrote: This problem was proposed by me. This problem was so nasty that I don't like it while it is chosen for the shortlist. My original solution uses an appropriate polynomial. I guess the appropriate polynomial is $P(x)=T(x-a)(x-c)(x-e)+S(x-b)(x-d)(x-f)$, clearly positive for $x=b,c,f$, and negative for $x=a,d,e$, hence it has exactly three real roots $u,v,w$ such that $u\in(a,b)$, $v\in(c,d)$ and $w\in(e,f)$, and now we have two options: 1) Using (very little) calculus, by Rolle's theorem $P'(x)$ must be zero both in $(u,v)$ and in $(v,w)$, or $P'(x)=3(S+T)x^2-4STx+(ac+ce+ea)T+(bd+df+fb)S$ has two distinct real roots, hence its discriminant is positive, which is equivalent to the proposed result. 2) Not using calculus (more "Olympic" solution), one can say that $u,v,w$ are distinct positive reals such that $u+v+w=\frac{2ST}{S+T}$ and $uv+vw+wu=\frac{T(ac+ce+ea)+S(bd+df+fb)}{S+T}$, and it suffices to apply Cauchy-Schwarz or AM-QM to find that $(u+v+w)^2>3(uv+vw+wu)$, equality not being possible since $u,v,w$ are distinct. Direct substitution yields the proposed result.

orl wrote: Given six positive numbers $a,b,c,d,e,f$ such that $a < b < c < d < e < f.$ Let $a+c+e=S$ and $b+d+f=T.$ Prove that \[2ST > \sqrt{3(S+T)\left(S(bd + bf + df) + T(ac + ae + ce) \right)}.\] Proposed by Sung Yun Kim, South Korea I have just seen this problem today. It is very interesting, so I will post my solution. Indeed, a Cauchy-Schwarz solution.

Squaring both sides, we can write the inequality as \[4S^2T^2 >3(S+T)\left[ S(bd+df+fb)+T(ac+ce+ea)\right],\] or \[\frac{4ST}{S+T} >\frac{3(bd+df+fb)}{T}+\frac{3(ac+ce+ea)}{S}.\] This is equivalent to \[\left[ T-\frac{3(bd+df+fb)}{T}\right] +\left[ S-\frac{3(ac+ce+ea)}{S}\right] \ge S+T-\frac{4ST}{S+T},\] or \[\frac{(f-b)^2+(f-d)^2+(d-b)^2}{2T}+\frac{(e-a)^2+(c-a)^2+(e-c)^2}{2S} >\frac{(T-S)^2}{S+T}.\] Click to reveal hidden text

Now, using the Cauchy-Schwarz inequality, we have \[\begin{aligned}\frac{(f-b)^2+(f-d)^2}{2T}+\frac{(e-a)^2+(c-a)^2}{2S} &\ge \frac{(2f-b-d)^2}{4T}+\frac{(e+c-2a)^2}{4S} \\ &\ge \frac{(2f+e+c-b-d-2a)^2}{4(S+T)}.\end{aligned}\] On the other hand, we also have $\frac{(d-b)^2}{2T}+\frac{(e-c)^2}{2S} >0.$ Therefore, it suffices to prove that \[\frac{(2f+e+c-b-d-2a)^2}{4(S+T)} \ge \frac{(f+b+d-e-c-a)^2}{S+T},\] or (notice that $2f+e+c-b-d-2a>0$ and $f+b+d-e-c-a>0$) \[2f+e+c-b-d-2a \ge 2(f+b+d-e-c-a).\] This inequality can be simplified to \[3(e+c) \ge 3(b+d),\] which is obviously true. The proof is completed. $\blacksquare$ Sung-yoon Kim wrote: This problem was proposed by me. This problem was so nasty that I don't like it while it is chosen for the shortlist. My original solution uses an appropriate polynomial. I think it is good enough.

$\Delta(a, b) :=AM(a, b)-GM(a, b)$. It can suffice to prove that $2\Delta(S^3T, ST^3)<(S+T)(S\sum_{cyc}\Delta(a^2, c^2)+T\sum_{cyc}\Delta(b^2, d^2))$. Fix $S$ and $T$. We know that minimum value of right hand is $T(S+T)(T-S)^2$ and this value is attained if and only if $a=b=c=d=e=\frac{S}{3}$ and $f=T-\frac{2S}{3}$. So it suffice to prove that $ST(T-S)^2<(S+T)T(T-S)^2$, and it is obvious.

Beautiful problem. Define $p(x)=T(x-a)(x-c)(x-e)$ and $q(x)=S(x-b)(x-d)(x-f)$. Let $\phi=p+q$. Observe that by the intermediate value theorem, $\phi$ has three real roots by the intermediate value theorem, one in the interval $(a,b)$, another in the interval $(c,d)$, and a third in the interval $(e,f)$. We can expand $\phi$ as \[(S+T)x^3-2STx^2+[T(ac+ce+ea)+S(bd+df+fb)]x-Tace-Sbdf.\]If the roots of this polynomial are $r,s,t$, then $r+s+t=\frac{2ST}{S+T}$ and \[rs+st+tr=\frac{T(ac+ce+ea)+S(bd+df+fb)}{S+T}.\]It is clear that $rs+st+tr\le \frac 13 (r+s+t)^2$ (and equality can't actually hold because $r,s,t$ must be distinct). This rearranges to \[(2ST)^2> 3(S+T)[T(ac+ce+ea)+S(bd+df+fb)],\]which is the desired.

orl wrote: Given six positive numbers $a,b,c,d,e,f$ such that $a < b < c < d < e < f.$ Let $a+c+e=S$ and $b+d+f=T.$ Prove that \[2ST > \sqrt{3(S+T)\left(S(bd + bf + df) + T(ac + ae + ce) \right)}.\] Proposed by Sung Yun Kim, South Korea Let $P(x) = T(x-a)(x-c)(x-e) + S(x-b)(x-d)(x-f)$. We see that $T > S$. Therefore, we see that due to Intermediate Value Theorem, there exists roots of $P$ in the intervals $(a, b), (c, d)$ and $(e, f)$, let these roots be $x, y, z$. We see that $P(x) = x^3(S +T) - 2x^2ST + x(T(ac + ce + ea) + S(bd + df + fb)) - (Tace + Sbdf)$. We see that since $x, y, z$ are roots of the polynomial $P$, we have $x + y + z = \dfrac{2ST}{S + T}$ and $xy + yz + zx = \dfrac{T(ac + ce + ea) + S(bd + df + fb)}{S + T}$ due to Vieta's relations. We see that since $x, y$ and $z$ are all distinct, $(x+y+z)^2 \geq 3(xy+yz+zx)$ (equality happens only at $x = y = z$). This means that $\dfrac{(2ST)^2}{(S+T)^2} = (x+y+z)^2 > 3(xy+yz+zx) = \dfrac{3T(ac + ce + ea) + 3S(bd + df + fb)}{S + T}$. This means that $(2ST)^2 > 3(S+T)(T(ac + ce + ea) + S(bd + df + fb)) \implies 2ST > \sqrt{3(S+T)(T(ac + ce + ea) + S(bd + df + fb)) }$.

hi can any one help me with this problem i ve seen seen the aops solution but didnt understand we have 0<=x,y,z<=1 prove that (x/(y+z+1))+(y/(x+z+1))+(z/(x+y+1))+(1-x)(1-y)(1-z)<=1

Let $A=ac+ce+ea=\tfrac{2S^2-C}{6}$, $B=bd+df+fb=\tfrac{2T^2-D}{6}$, then \begin{align*} C=2S^2-6A &= (a-c)^2+(c-e)^2+(e-a)^2 \\ D=2T^2-6B &= (b-d)^2+(d-f)^2+(f-b)^2 \end{align*}and we want to show $2ST>\sqrt{(S+T)(3SB+3TA)}$. Since $ST$ and $\sqrt{(S+T)(3SB+3TA)}$ are both positive, this is equivalent to showing \begin{align*} 8S^2T^2 &> (S+T)(S(6B) + T(6A)) \\ &= (S+T)(S(2T^2-D)+T(2S^2-C)) \\ &= 4S^2T^2 + 2ST^3 + 2S^3T - (S+T)(SD+TC) \end{align*}Rearranging, we obtain the equivalent statement $(S+T)(SD+TC)>2ST(S-T)^2$. By C-S, \[(S+T)(SD+TC)\ge (\sqrt{STC}+\sqrt{STD})^2=ST(\sqrt{C}+\sqrt{D})^2\]So it suffices to show $(\sqrt{C}+\sqrt{D})^2>2(S-T)^2$. Since $T-S$ and $\sqrt{C}+\sqrt{D}$ are both positive, this is equivalent to \[\sqrt{\frac{C}{2}}+\sqrt{\frac{D}{2}}>T-S\]By QM-AM, \begin{align*} \sqrt{\frac{C}{2}}+\sqrt{\frac{D}{2}} &= \sqrt{\frac{(a-c)^2+(c-e)^2+(e-a)^2}{2}}+\sqrt{\frac{(b-d)^2+(d-f)^2+(f-b)^2}{2}}\\ &> \sqrt{\frac{(a-c)^2+(e-a)^2}{2}}+\sqrt{\frac{(d-f)^2+(f-b)^2}{2}} \\ &> \frac{e+c-2a}{2}+\frac{2f-b-d}{2} \\ &= (f-a) + \frac{e+c-b-d}{2} \\ &> (f-a) + \frac{-2e-2c+2b+2d}{2} \\ &= T-S \end{align*}as desired.

omgomgomg A8 inequalities are not impossible!! The following solution feels more natural and "safer" than the polynomial solution; its only non-equality steps are a two-term Cauchy-Schwarz in the middle and a few dropped terms in the end. Let $U=ac+ce+ea$ and $V=bd+df+fb$. It suffices to prove \[4S^2T^2>3(S+T)(SV+TU) \Longleftrightarrow 4ST>3(S+T)\left(\frac{U}{S}+\frac{V}{T}\right).\]We have \[\frac{U}{S}=\frac{\frac{S^2}{3}-\frac{a^2+c^2+e^2-ac-ce-ea}{3}}{S}=\frac{S}{3}-\frac{(c-a)^2+(e-c)^2+(e-a)^2}{6S}\]and \[\frac{V}{T}=\frac{\frac{T^2}{3}-\frac{b^2+d^2+f^2-bd-df-fb}{3}}{T}=\frac{T}{3}-\frac{(d-b)^2+(f-d)^2+(f-b)^2}{6T},\]so it suffices to prove \begin{align*} &4ST>(S+T)\left(S+T-\frac{(c-a)^2+(e-c)^2+(e-a)^2}{2S}-\frac{(d-b)^2+(f-d)^2+(f-b)^2}{2T}\right) \\ \Longleftrightarrow \ & 4ST>(S+T)^2-(S+T)\left(\frac{(c-a)^2+(e-c)^2+(e-a)^2}{2S}+\frac{(d-b)^2+(f-d)^2+(f-b)^2}{2T}\right) \\ \Longleftrightarrow \ & (S+T)\left(\frac{(c-a)^2+(e-c)^2+(e-a)^2}{2S}+\frac{(d-b)^2+(f-d)^2+(f-b)^2}{2T}\right)>(T-S)^2 \end{align*}By Cauchy-Schwarz, it suffices to prove \begin{align*} &\left(\sqrt{\frac{(c-a)^2+(e-c)^2+(e-a)^2}{2}}+\sqrt{\frac{(d-b)^2+(f-d)^2+(f-b)^2}{2}}\right)^2>(T-S)^2 \\ \Longleftrightarrow \ & \sqrt{\frac{(c-a)^2+(e-c)^2+(e-a)^2}{2}}+\sqrt{\frac{(d-b)^2+(f-d)^2+(f-b)^2}{2}}>T-S. \end{align*}Let $(p,q,r,s,t):=(b-a,c-b,d-c,e-d,f-e)$. Then, it suffices to prove \[\sqrt{\frac{(p+q)^2+(r+s)^2+(p+q+r+s)^2}{2}}+\sqrt{\frac{(q+r)^2+(s+t)^2+(q+r+s+t)^2}{2}}>p+r+t.\]Since $q$ and $r$ contribute positively to the LHS, it suffices to prove \[p+r+t \le \sqrt{\frac{p^2+r^2+(p+r)^2}{2}}+\sqrt{\frac{r^2+t^2+(r+t)^2}{2}}=\sqrt{p^2+pr+r^2}+\sqrt{t^2+tr+r^2}.\]But we have \[\sqrt{p^2+pr+r^2}+\sqrt{t^2+tr+r^2}>\sqrt{p^2+pr+\frac{r^2}{4}}+\sqrt{t^2+tr+\frac{r^2}{4}}=p+\frac{r}{2}+t+\frac{r}{2}=p+r+t,\]as desired. $\square$