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

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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 $\prod_{cyc}(49x-7y+z)=(49x-7y+z)(49y-7z+x)(49x-7y+z)$