Olympiad inequality $\Big(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\Big)\sqrt{\frac{\prod_{cyc}(49x-7y+z)}{43^3}}\leq \sqrt{3(x+y+z)}$

I'm interested by the following problem :

Let $$x,y,z>0$$ with $$49x-7y+z>0$$, $$49y-7z+x>0$$ , $$49z-7x+y>0$$ then we have : $$\Big(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\Big)\sqrt{\frac{\prod_{cyc}(49x-7y+z)}{43^3}}\leq \sqrt{3(x+y+z)}$$

I have tested this inequality with Pari-Gp and it seems to be okay . Furthermore I think we can use the $$uvw$$'s method (because the equality case comes when $$x=y=z$$) but I don't see how now . I have a ugly proof using derivative (the inequality can be reduce to a two variable inequality) but it's too long to be explain here . The inequality is too precise to use Jensen's or Slater's inequality here. Finally I have also tested brut force but I can't find an interesting irreductible factorization .

If you have a hint it would be nice .

Thanks in advance .

Edit :With the notation we have $$\prod_{cyc}(49x-7y+z)=(49x-7y+z)(49y-7z+x)(49x-7y+z)$$

• What is the $\Pi_{cyc}$ notation? – Calvin Godfrey May 31 at 12:33
• You can square both sides and then try the SOS method. – Dr. Sonnhard Graubner May 31 at 12:39
• SOS means Sum Of Squares – Dr. Sonnhard Graubner May 31 at 12:40
• From where does this problem come? – Dr. Sonnhard Graubner May 31 at 12:41
• @CalvinGodfrey see this – user574848 May 31 at 13:07