It looks terribly hard at a glance. It turns out to be an old problem of Vietnamese Mathematical Olympiad 1996. After simple substitution $a=\sqrt{\frac{vw}u}$..., we can write it as Let $a,b,c$ be positive reals such that \[ab+bc+ca+abc=4, \] Prove that $a+b+c\geq ab+bc+ca.$ Is it an evidence that they are runing out of new inequalities for national Olympiads? If at all, what a pity!

It is a problem from VietnamMO @shencaili: Can you post all problems from China TST 2007 to National Olympiad box?

I'm sorry,but we've just had two exams.

Otherwise, it is the intention of the proposer to remind his Chinese team members of Vietnam as a host country for this coming IMO. All for a memorable IMO in Vietnam

is this proof correct: :we have to prove $z+x+y\ge zx+xy+yz$ when $x+z+y+xyz=4$ let$x\ge y\ge z$ so that $z\le 1$ \[x+z+y+xzy=4 \ \iff x=\frac{4-y-z}{1+yz}\] substituting $x$ we have to prove \[\frac{4-y-z}{1+yz}+y+z\ge \frac{(4-y-z)}{1+yz}.(y+z)+yz\] \[\iff 4+yz(y+z)\ge 4(y+z)-(y+z)^{2}+yz+y^{2}z^{2}\] it becomes to prove $f(y)=(1+z-z^{2})y^{2}+y(z^{2}-4+z)+(4-4z+z^{2})\ge 0$ which is equivalent to it having nonpositive discriminent \[D=(z^{2}+z-4)^{2}-4(4-4z+z^{2})(z+1-z^{2})\le 0\] \[\iff (z-1)^{2}(5z-8)\le 0\] which is true as $z\le 1$ by assumption.

Yes, that is perfectly right.

thanks pvthuan!

anuj kumar wrote: is this proof correct: :we have to prove $z+x+y\ge zx+xy+yz$ when $x+z+y+xyz=4$ let$x\ge y\ge z$ so that $z\le 1$ \[x+z+y+xzy=4 \ \iff x=\frac{4-y-z}{1+yz}\] substituting $x$ we have to prove \[\frac{4-y-z}{1+yz}+y+z\ge \frac{(4-y-z)}{1+yz}.(y+z)+yz \] \[\iff 4+yz(y+z)\ge 4(y+z)-(y+z)^{2}+yz+y^{2}z^{2}\] it becomes to prove $f(y)=(1+z-z^{2})y^{2}+y(z^{2}-4+z)+(4-4z+z^{2})\ge 0$ which is equivalent to it having nonpositive discriminent \[D=(z^{2}+z-4)^{2}-4(4-4z+z^{2})(z+1-z^{2})\le 0 \] \[\iff (z-1)^{2}(5z-8)\le 0 \] which is true as $z\le 1$ by assumption. but you made a mistake.Let be positive reals such that Prove that

$u,v,w>0$,such that $u+v+w+\sqrt{uvw}=4$ prove that $\sqrt{\frac{uv}{w}}+\sqrt{\frac{vw}{u}}+\sqrt{\frac{wu}{v}}\geq u+v+w$ ............(1) proof:Let \[a^{2}= \frac{1}{4}u,b^{2}= \frac{1}{4}v,c^{2}= \frac{1}{4}w,{\rm{ }}{\rm{ }}a,b,c > 0 \] we can write it as Let be positive reals such that $a^{2}+b^{2}+c^{2}+abc = 1$...........(2) Prove that $\frac{{bc}}{a}+\frac{{ca}}{b}+\frac{{ab}}{c}\ge 2(a^{2}+b^{2}+c^{2})$ ........... (3) by (2),Let$a = \cos A,b = \cos B,c = \cos C$ thereinto $A,B,C$ are the angles of acute angle$\Delta ABC$, so that $(3) \Leftrightarrow \frac{{\cos B\cos C}}{{\cos A}}+\frac{{\cos C\cos A}}{{\cos B}}+\frac{{\cos A\cos B}}{{\cos C}}\ge 2(\cos^{2}A+\cos^{2}B+\cos^{2}C)$ ...... (4) By"$x^{2}+y^{2}+z^{2}\ge 2yz\cos A+2zx\cos B+2xy\cos C$,thereinto $A,B,C$ are the angles of $\Delta ABC$,$x,y,z \in R$“，we have $\sum_{CycA,B,C}{\frac{{\cos B\cos C}}{{\cos A}}}\ge 2\sum_{CycA,B,C}{\sqrt{\frac{{\cos C\cos A}}{{\cos B}}}}\cdot \sqrt{\frac{{\cos A\cos B}}{{\cos C}}}\cdot \cos A = 2\sum_{CycA,B,C}{\cos^{2}A}$

Another one: $a_{1},...,a_{n}>0$ that $a_{1}+..+a_{n}=1$ prove that $(a_{1}a_{2}+..+a_{n}a_{1})(\frac{a_{1}}{a_{2}^{2}+a_{2}}+...)\geq\frac{n}{n+1}$

another one: $a,b,c \in (0,1], a^{2}+b^{2}+c^{2}=2$ prove that $\frac{1-b^{2}}{a}+\frac{1-c^{2}}{b}+\frac{1-a^{2}}{c}\leq \frac{5}{4}$ i think it is much harder than the first one and the second one, we can use schur inequality to solve the first one

The second inequality if $\sum_{i=1}^{n}a_{i}a_{i+1}\geq \frac{1}{n}$ $\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}(a_{i+1}+1)}\sum_{i=1}^{n}(a_{i+1}+1)\geq (\sum_{i=1}^{n}(\frac{a_{i}}{a_{i+1}})^\frac{1}{2})^{2}\geq n^{2}$ $\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}(a_{i+1}+1)}\geq \frac{n^{2}}{n+1}$ then $\sum_{i=1}^{n}a_{i}a_{i+1}\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}(a_{i+1}+1)}\geq \frac{n}{n+1}$ if $\sum_{i=1}^{n}a_{i}a_{i+1}<\frac{1}{n}$ $(\sum_{i=1}^{n}a_{i}a_{i+1})(\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}(a_{i+1}+1)})(\sum_{i=1}^{n}a_{i}(a_{i+1}+1))\geq (\sum_{i=1}^{n}a_{i})^{3}=1$ And $\sum_{i=1}^{n}a_{i}(a_{i+1}+1)<\frac{n+1}{n}$ We have $(\sum_{i=1}^{n}a_{i}a_{i+1})(\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}(a_{i+1}+1)})>\frac{n}{n+1}$ BINGO!

anuj kumar wrote: is this proof correct: :we have to prove $z+x+y\ge zx+xy+yz$ when $x+z+y+xyz=4$ let$x\ge y\ge z$ so that $z\le 1$ \[x+z+y+xzy=4 \ \iff x=\frac{4-y-z}{1+yz}\] substituting $x$ we have to prove \[\frac{4-y-z}{1+yz}+y+z\ge \frac{(4-y-z)}{1+yz}.(y+z)+yz \] \[\iff 4+yz(y+z)\ge 4(y+z)-(y+z)^{2}+yz+y^{2}z^{2}\] it becomes to prove $f(y)=(1+z-z^{2})y^{2}+y(z^{2}-4+z)+(4-4z+z^{2})\ge 0$ which is equivalent to it having nonpositive discriminent \[D=(z^{2}+z-4)^{2}-4(4-4z+z^{2})(z+1-z^{2})\le 0 \] \[\iff (z-1)^{2}(5z-8)\le 0 \] which is true as $z\le 1$ by assumption. I think it is not right as you assume $x+z+y+xyz=4$ though it should be $xy+xz+yz+xyz=4$ @xuheiedgar: Can you please show how to prove the inequality with Schur? Thank you.

I'm a friend of xuheiedgar. He and I use the same method to solve this problem in the exam (Schur). He's not at home now so I'll post the solution. Take $a=\sqrt{\frac{vw}{u}},\ b=\sqrt{\frac{wu}{v}},\ c=\sqrt{\frac{uv}{w}}$ we have $ab+bc+ca+abc=4$ We just need to prove $a+b+c\geqslant ab+bc+ca$ Actually if $a+b+c<ab+bc+ca$ With Schur we have $a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\geqslant0\\ \Longleftrightarrow(a+b+c)^{3}+9abc\geqslant 4(a+b+c)(ab+bc+ca)$ So $9abc\geqslant[4(ab+bc+ca)-(a+b+c)^{2}](a+b+c)\\ >[4(ab+bc+ca)-(ab+bc+ca)^{2}](a+b+c)\\ =abc(a+b+c)(ab+bc+ca)$ So we have $\sqrt{3(ab+bc+ca)}\cdot(ab+bc+ca)\leqslant(a+b+c)(ab+bc+ca)<9$ That means $ab+bc+ca<3$ Then it's easy that $abc\leqslant\left(\frac{ab+bc+ca}{3}\right)^{3/2}<1$ Hence $ab+bc+ca+abc<4$ Till now it's done.

thank you guys for pointing out my mistake .actually i had seen it wrongly .but same method works in correct condition too and to surprise gives the same quadratic function so we have to prove $x+y+z\ge zx+xy+yz$ when $xz+zy+yx+xyz=4$ again from condition we have $x=\frac{4-yz}{z+y+yz}$ and which on substitution wants us to prove \[\frac{4-yz}{z+y+zy}+y+z\ge \frac{4-yz}{z+y+zy}.(y+z)+zy\] \[\iff y^{2}(1+z-z^{2})+y(z+z^{2}-4)+(4-4z+z^{2})\ge 0 \iff D=(z-1)^{2}.(5z-8)\le 0\] which is true. nevertheless my mistake gave rise to newer inequality.

shencaili wrote: $u,v,w>0$,such that $u+v+w+\sqrt{uvw}=4$ prove that $\sqrt{\frac{uv}{w}}+\sqrt{\frac{vw}{u}}+\sqrt{\frac{wu}{v}}\geq u+v+w$ $u+v+w+\sqrt{uvw}=4\Longleftrightarrow \sqrt{uvw}=4-(u+v+w)\ (u,v,w>0).$ Thus we can rewrite our inequality as follows. $\sqrt{\frac{uv}{w}}+\sqrt{\frac{vw}{u}}+\sqrt{\frac{wu}{v}}\geq u+v+w$ $\Longleftrightarrow uv+vw+wu\geq (u+v+w)\sqrt{uvw}$ $\Longleftrightarrow uv+vw+wu\geq (u+v+w)\{4-(u+v+w)\}$ $\Longleftrightarrow(u+v+w)^{2}-4(u+v+w)+uv+vw+wu\geq 0$ $\Longleftrightarrow (u+v+w-2)^{2}+uv+vw+wu-4\geq 0.$ Under the condition $0<u+v+w<4,$ We are to prove that $uv+vw+wu\geq 4.$

shencaili wrote: $ u,v,w > 0$,such that $ u + v + w + \sqrt {uvw} = 4$ prove that $ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w$ Let $ u = \frac {4yz}{(x + y)(x + z)},$ $ v = \frac {4xz}{(x + y)(y + z)},$ where $ x,$ $ y$ and $ z$ are positive numbers. Hence, $ w = \frac {4xy}{(x + z)(y + z)}$ and $ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w\Leftrightarrow uv + uw + vw\geq(u + v + w)\sqrt {uvw}\Leftrightarrow$ $ \Leftrightarrow\sum_{cyc}\frac {16z^2xy}{(x + y)^2(x + z)(y + z)}\geq\sum_{cyc}\frac {4xy}{(x + z)(y + z)}\cdot\frac {8xyz}{(x + y)(x + z)(y + z)}\Leftrightarrow$ $ \Leftrightarrow\sum_{cyc}(x^3 - x^2y - x^2z + xyz)\geq0,$ which is Schur.

from titu andrescu 'problems from the books' $ab+ac+bc+abc=4$. Denote m,n,k for which $\frac{2}{m}=a,\frac{2}{n}=b,\frac{2}{k}=c => mnk=m+n+k+2 => \frac{1}{m+1}+\frac{1}{n+1}+\frac{1}{k+1}=1$ $x=\frac{1}{m+1},y=\frac{1}{n+1},z=\frac{1}{k+1}=>m=\frac{y+z}{x},n=\frac{x+z}{y},k=\frac{y+z}{x}=>a=\frac{2x}{y+z},b=\frac{2y}{x+z},c=\frac{2z}{x+y} $ and we want prove that $a+b+c=\frac{2x}{y+z}+\frac{2y}{x+z}+\frac{2z}{x+y}\geq ab+ac+bc=\frac{2x}{y+z}.\frac{2y}{x+z}+\frac{2x}{y+z}.\frac{2z}{x+y}+\frac{2y}{x+z}.\frac{2z}{x+y}$ This inequality equivalent to $\sum_{cyc}x^{3}+\sum_{sym}x^{2}y+3xyz\geq 2\sum_{sym}x^{2}y$ and this is just schur

Here is another solution: We may rewrite the given condition as $\left(\sqrt{u}\right)^2 + \left(\sqrt{v}\right)^2 + \left(\sqrt{w}\right)^2 + \left(\sqrt{u}\right)\left(\sqrt{v}\right)\left(\sqrt{w}\right) = 4.$ It is then well-known that the triple $\left(\sqrt{u}, \sqrt{v}, \sqrt{w}\right)$ may be represented as $(2\cos A, 2\cos B, 2\cos C)$ where $A, B, C$ are the angles of an acute triangle. This follows from the well-known identity $\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A\cos B\cos C = 1$; in particular, since $\max\{\sqrt{u}, \sqrt{v}, \sqrt{w}\} < 2$, we can set $\sqrt{u} = 2\cos A, \sqrt{v} = 2\cos B$ for angles $A, B \in \left(0, \frac{\pi}{2}\right)$, whence solving the resulting quadratic equation in $\sqrt{w}$ yields $\sqrt{w} = -2\cos(A + B) = 2\cos C.$

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shencaili wrote: $ u,v,w > 0$,such that $ u + v + w + \sqrt {uvw} = 4$ prove that $ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w$ let $u=x^2, v=y^2,w=z^2 $ hence $x^2+y^2+z^2+xyz=4 $ $\iff x^2y^2+y^2z^2+z^2x^2\geq xyz(x^2+y^2+z^2) $ let $ x=2\sqrt{\frac{bc}{(a+b)(a+c)}} ,y=2\sqrt{\frac{ac}{(a+b)(b+c)}}, x=2\sqrt{\frac{ba}{(a+c)(b+c)}} $ $\iff \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq 2(\frac{ab}{(a+c)(b+c)}+\frac{bc}{(a+c)(b+a)}+\frac{ac}{(a+b)(b+c)}) $ $\iff a^3+b^3+c^3+3abc\geq a^2b+b^2a+c^2a+a^2c+b^2c+c^2b $ this is true.

Substitute $x^2=u, y^2=v, z^2=w$. So we have $x^2+y^2+z^2+xyz=4$. Thus by the identity $\cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1$, we can write $x=2\cos A, y = 2 \cos B, z = 2 \cos C$ where $A, B, C$ are three angles of an acute triangle. So it remains to prove that \[ \sum_{cyc} \frac{\cos A \cos B}{\cos C} \ge 2 \cos^2 A + 2 \cos^2 B + 2 \cos^2 C\]Applying the lemma \[ x^2+y^2+z^2 \ge 2xy \cos B + 2yz \cos C + 2zx \cos A\]with $x^2=\frac{\cos A \cos B}{\cos C}, y^2= \frac{\cos B \cos C}{\cos A}, z^2 = \frac{\cos C \cos A}{\cos B}$ yields the desired. By the way, I was trying to prove the inequality $\sum_{cyc} \frac{\cos A \cos B}{\cos C} \ge 2 \cos^2 A + 2 \cos^2 B + 2 \cos^2 C$ with the quadratic formula after the identity $\cos^2 A +\cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C=1$ (treating as quadratic in $\cos A \cos B \cos C$) or using Cauchy. However I couldn't do so - can anyone find a solution with the two aforementioned methods?

shencaili wrote: $ u,v,w > 0$,such that $ u + v + w + \sqrt {uvw} = 4$ prove that $ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w$ Let $z=\sqrt{\frac{uv}{w}},x=\sqrt{\frac{vw}{u}},y=\sqrt{\frac{wu}{v}}.$ We need to prove that if $xy+yz+zx+xyz=4$, then $x+y+z\ge xy+yz+zx$. It's equivalent to prove that if $x+y+z=xy+yz+zx$, then $xy+yz+zx+xyz\ge 4.$ Let $p=x+y+z,q=xy+yz+zx,r=xyz.$ Note that $p=q$, so $q+r\ge 4 \iff q\frac{q}{p}+r\ge 4\frac{q^3}{p^3}\iff p^2q^2+p^3r\ge 4q^3.$ By Schur we have $p^3-4pq+9r\ge 0\implies p^2q^2+\frac{9q^2r}{p}\ge 4q^3.$ We just need to prove $p^3r\ge \frac{9q^2r}{p}$, which is trivial since $p^2\ge 3q.\blacksquare$

$\Delta ABC$ is an acute triangle. prove that: $$\cos A\sin B+\cos B\sin C+\cos C\sin A<\dfrac{5\sqrt{2(\sin^2 A+\sin^2 B+\sin^2 C)}}{8}$$(2007 China TST ?)

shencaili wrote: $ u,v,w > 0$,such that $ u + v + w + \sqrt {uvw} = 4$ prove that $ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w$ Beautiful

shencaili wrote: $ u,v,w > 0$,such that $ u + v + w + \sqrt {uvw} = 4$ prove that $$ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w$$ Let $ u, v, w$ be positive real numbers such that $ u\sqrt {vw} + v\sqrt {wu} + w\sqrt {uv} \geq 1$. Prove that$$ u + v + w\geq \sqrt 3$$

Scrutiny wrote: shencaili wrote: $ u,v,w > 0$,such that $ u + v + w + \sqrt {uvw} = 4$ prove that $ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w$ Let $z=\sqrt{\frac{uv}{w}},x=\sqrt{\frac{vw}{u}},y=\sqrt{\frac{wu}{v}}.$ We need to prove that if $xy+yz+zx+xyz=4$, then $x+y+z\ge xy+yz+zx$. It's equivalent to prove that if $x+y+z=xy+yz+zx$, then $xy+yz+zx+xyz\ge 4.$ Let $p=x+y+z,q=xy+yz+zx,r=xyz.$ Note that $p=q$, so $q+r\ge 4 \iff q\frac{q}{p}+r\ge 4\frac{q^3}{p^3}\iff p^2q^2+p^3r\ge 4q^3.$ By Schur we have $p^3-4pq+9r\ge 0\implies p^2q^2+\frac{9q^2r}{p}\ge 4q^3.$ We just need to prove $p^3r\ge \frac{9q^2r}{p}$, which is trivial since $p^2\ge 3q.\blacksquare$ inventive

Let $ u,v,w > 0$ and $u + v + w + \sqrt {uvw}=4 .$ Prove that $$ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w \geq \sqrt {u} + \sqrt {v} + \sqrt {w}$$

AMN300 wrote: Substitute $x^2=u, y^2=v, z^2=w$. So we have $x^2+y^2+z^2+xyz=4$. Thus by the identity $\cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1$, we can write $x=2\cos A, y = 2 \cos B, z = 2 \cos C$ where $A, B, C$ are three angles of an acute triangle. So it remains to prove that \[ \sum_{cyc} \frac{\cos A \cos B}{\cos C} \ge 2 \cos^2 A + 2 \cos^2 B + 2 \cos^2 C\]Applying the lemma \[ x^2+y^2+z^2 \ge 2xy \cos B + 2yz \cos C + 2zx \cos A\]with $x^2=\frac{\cos A \cos B}{\cos C}, y^2= \frac{\cos B \cos C}{\cos A}, z^2 = \frac{\cos C \cos A}{\cos B}$ yields the desired. By the way, I was trying to prove the inequality $\sum_{cyc} \frac{\cos A \cos B}{\cos C} \ge 2 \cos^2 A + 2 \cos^2 B + 2 \cos^2 C$ with the quadratic formula after the identity $\cos^2 A +\cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C=1$ (treating as quadratic in $\cos A \cos B \cos C$) or using Cauchy. However I couldn't do so - can anyone find a solution with the two aforementioned methods? Super nice

shencaili wrote: $ u,v,w > 0$,such that $ u + v + w + \sqrt {uvw} = 4$ prove that $ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w$

Let $u=4\cos ^2A,v=4\cos ^2B,w=4\cos ^2C.$ Then $$u+v+w+\sqrt{uvw}=4(\cos ^2+\cos ^2B+\cos ^2C+2\cos A\cos B\cos C)=4.$$$$ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geqslant u + v + w$$$$\Leftrightarrow\sum_{cyc}\frac{2\cos A\cos B}{\cos C}\geqslant 4\sum_{cyc}\cos ^2A.$$$$\Leftrightarrow\sum_{cyc}\left(\sqrt{\frac{\cos A\cos B}{\cos C}}\right)^2\geqslant 2\sum_{cyc}\sqrt{\frac{\cos B\cos C}{\cos A}}\cdot\sqrt{\frac{\cos C\cos A}{\cos B}}\cdot\cos A.$$This is the 嵌入不等式.$\blacksquare$