If x,y,z are positive real numbers and xy+yz+zx=1 prove that 274(x+y)(y+z)(z+x)≥(√x+y+√y+z+√z+x)2≥6√3.
Problem
Source: Turkey, TST D1, P3
Tags: inequalities, inequalities proposed, algebra
10.05.2006 18:17
The inequality 9(x+y)(y+z)(z+x)≥8(xy+yz+zx)(x+y+z) (1) it seems to be lovely Using (1) and after Cauchy We have 274(x+y)(y+z)(z+x)≥6(x+y+z)≥(√x+y+√y+z+√z+x)2 Now by using AM-GM and then (1) we have (√x+y+√y+z+√z+x)2≥93√(x+y)(y+z)(z+x)≥93√8(x+y+z)9≥93√8√3(xy+yz+zx)9≥6√3 and we are done
11.05.2006 11:05
The inequality is very easy if we substitute x=tanA2,y=tanB2,z=tanc2 where A,B,C are angles of the triangle. Then we get easy inequalities(very typical) that are followed by Jensen.
12.05.2006 07:31
Looks so familiar. I guess I posted my solution too early in the wrong thread.
02.10.2006 00:23
substitute by cotga, cotgb, cotgc so that a+b+c=180° in a acute triangle and use jensen
17.08.2009 23:06
Can someone explain the Jensen solution fully? I tried the cotangent trick but couldn't get it to work.
14.05.2019 06:31
Valentin Vornicu wrote: If x,y,z are positive real numbers and xy+yz+zx=1 prove that 274(x+y)(y+z)(z+x)≥(√x+y+√y+z+√z+x)2≥6√3. But you do this stronger: If x,y,z are positive real numbers and xy+yz+zx=1 prove that (2√2+3)(x+y)(y+z)(z+x)+8(154−2√2)xyz≥(√x+y+√y+z+√z+x)2
14.05.2019 06:37
xzlbq wrote: Valentin Vornicu wrote: If x,y,z are positive real numbers and xy+yz+zx=1 prove that 274(x+y)(y+z)(z+x)≥(√x+y+√y+z+√z+x)2≥6√3. But you do this stronger: If x,y,z are positive real numbers and xy+yz+zx=1 prove that (2√2+3)(x+y)(y+z)(z+x)+8(154−2√2)xyz≥(√x+y+√y+z+√z+x)2 By C-S,need to prove (2√2+3)(x+y)(y+z)(z+x)+8(154−2√2)xyz≥6x+6y+6z <=> xz2+yz2+x2y+xy2+x2z−6xyz+y2z≥0
14.05.2019 06:44
Valentin Vornicu wrote: If x,y,z are positive real numbers and xy+yz+zx=1 prove that 274(x+y)(y+z)(z+x)≥(√x+y+√y+z+√z+x)2≥6√3. Stronger is: If x,y,z are positive real numbers and xy+yz+zx=1 prove that (√x+y+√y+z+√z+x)2≥6√(x+y+z)√3
14.05.2019 06:48
xzlbq wrote: Valentin Vornicu wrote: If x,y,z are positive real numbers and xy+yz+zx=1 prove that 274(x+y)(y+z)(z+x)≥(√x+y+√y+z+√z+x)2≥6√3. Stronger is: If x,y,z are positive real numbers and xy+yz+zx=1 prove that (√x+y+√y+z+√z+x)2≥6√(x+y+z)√3 W=2z+x+y By Holder,need to prove (∑((x+y)(2z+x+y)))3∑((x+y)2(2z+x+y)3)≥6√(x+y+z)√3<=> (8(x+y+z)6+24(x+y+z)4+24(x+y+z)2+8)2(2(x+y+z)5+(x+y+z)3+3(x+y+z)2xyz+7x+7y+7z−5xyz)2≥36(x+y+z)√3 very hard!
28.09.2019 03:35
Let's first show the left inequality. After homogenizing, we want to show that: 27(x+y)(y+z)(z+x)≥4(xy+yz+zx)(√x+y+√y+z+√z+x)2. First of all, notice that 27(x+y)(y+z)(z+x)≥24(x+y+z)(xy+yz+zx) by AM-GM. Therefore, it suffices to show that 24(x+y+z)≥(√x+y+√y+z+√z+x)2. However, this is easy, because (√x+y+√y+z+√z+x)2≤3(x+y+y+z+z+x)=6(x+y+z). Let's now show the second inequality. We want to show that (√x+y+√y+z+√z+x)2≥6√3, whenever xy+yz+zx=1 and x,y,z are nonnegative reals. If any of x,y,z are bigger than 100, then we are clearly done. Else, let us restrict x,y,z to the interval [0,100]. Let's prove this with Lagrange Multipliers. The derivative of the LHS with respect to x is 2(√x+y+√y+z+√z+x)(12√x+y+12√x+z). The derivative of the condition w.r.t x is y+z. Therefore, we have that: 1√x+y+1√x+zy+z is cyclically constant. After multiplying by √(x+y)(y+z)(z+x), we get that: √x+y+√x+z√y+z is cyclically constant. This is easily seen to imply that x=y=z. Hence, the minimum is achieved either when xyz=0, max, or x = y = z = \frac{1}{\sqrt3}. It's easy to check that all of these cases are wins. \square
21.10.2021 18:00
Surprisingly simple. Let p = x + y + z , r = xyz. Note that p^2 \geq 3(xy + yz + zx) = 3 \implies p \geq \sqrt{3} and r \leq \frac{1}{3\sqrt{3}} by AM-GM. First we show the left part. It is equivalent to \frac{27}{4}(p - r) \geq 2p + 2\sum_{\text{cyc}} \sqrt{x^2 + 1}By C-S, \sum_{\text{cyc}} \sqrt{x^2 + 1} \leq \sqrt{3(x^2 + y^2 + z^2 + 3)} = \sqrt{3(p^2 + 1)}. So it suffices to show \frac{19}{4}p - 2\sqrt{3(p^2 + 1)} \geq \frac{27}{4}r. Note that the LHS is increasing in p and so \frac{19}{4}p - 2\sqrt{3(p^2 + 1)} \geq \frac{3\sqrt{3}}{4} \geq \frac{27}{4}r. Now let's show the right part. It is equivalent to 2p + 2\sum_{\text{cyc}} \sqrt{x^2 + 1} \geq 6\sqrt{3}. Since \sqrt{x^2 + 1} is convex, by Jensen we get 2\sum_{\text{cyc}} \sqrt{x^2 + 1} \geq 4\sqrt{3}. And since 2p \geq 2\sqrt{3}, we get the result. Done.