Let $a,$ $b$ and $c$ are positive numbers.

Question

Prove that $$\sqrt{\frac{a}{b+\alpha c}}+\sqrt{\frac{b}{c+\alpha a}}+\sqrt{\frac{c}{a+\alpha b}}\geq\frac{3}{\sqrt{1+\alpha}}$$ is true for all $\alpha\geq\dfrac{49+9\sqrt{17}}{32}$

I found this inequality proved in this way

let $x,y,z$ satisfies $\dfrac{a}{b+\alpha c}=\dfrac{x^2}{1+\alpha}$, $\dfrac{b}{c+\alpha a}=\dfrac{y^2}{1+\alpha}$, $\dfrac{c}{a+\alpha b}=\dfrac{z^2}{1+\alpha}$ we have $LHS=\dfrac{x+y+z}{\sqrt{1+\alpha}}$ and by using $$x^2y^2z^2=\dfrac{(1+\alpha)^3abc}{(a+\alpha b)(b+\alpha c)(c+\alpha a)}$$and $$x^2y^2+y^2z^2+z^2x^2=\frac{(1+\alpha)^2(ab(a+\alpha b)+bc(b+\alpha c)+ca(c+\alpha a)}{(a+\alpha b)(b+\alpha c)(c+\alpha a)}$$ we obtain$${(\alpha^2-\alpha+1)x^2y^2z^2+\alpha(x^2y^2+y^2z^2+z^2x^2)=(1+\alpha)^2}(*)$$ suppose that $x+y+z<3$ ,there is a $k$ satisfies $k(x+y+z)=3$ and $k>1$ let $p=kx,q=ky,r=kz$ and use $(*)$ we have $$(\alpha^2-\alpha+1)p^2q^2r^2+\alpha(p^2q^2+q^2r^2+r^2p^2)>(1+\alpha)^2$$ now we try to reach a contradiction by proving$$(\alpha^2-\alpha+1)p^2q^2r^2+\alpha(p^2q^2+q^2r^2+r^2p^2)\leq(1+\alpha)^2$$ for all $p,q,r,\ p+q+r=3$ which is equivalent to $$\alpha^2+\frac{2+p^2q^2r^2-(p^2q^2+q^2r^2+r^2p^2)}{1-p^2q^2r^2}\alpha+1\geq0(**)$$ but we have$$\frac{2+p^2q^2r^2-(p^2q^2+q^2r^2+r^2p^2)}{1-p^2q^2r^2}\geq\frac{49}{16}$$ since it can be rewritten as $$\sum{p^6}+6\sum{p^5(q+r)}-\sum{p^4(q^2+r^2)}-12\sum{p^3q^3}+30pqr\sum{p^3}+28pqr\sum{p^2(q+r)}-255p^2q^2r^2\geq0$$ which is schur and muirhead. so by using quadratic polynomial $(**)$ is true when $$\alpha\geq\frac{49+9\sqrt{17}}{32}$$ thus our assumption is wrong and $x+y+z\geq3$, $LHS\geq\dfrac{3}{\sqrt{1+\alpha}}$.Done.

Here the step

$$\alpha^2+\frac{2+p^2q^2r^2-(p^2q^2+q^2r^2+r^2p^2)}{1-p^2q^2r^2}\alpha+1\geq0(**)$$ but we have$$\frac{2+p^2q^2r^2-(p^2q^2+q^2r^2+r^2p^2)}{1-p^2q^2r^2}\geq\frac{49}{16}$$ since it can be rewritten as $$\sum{p^6}+6\sum{p^5(q+r)}-\sum{p^4(q^2+r^2)}-12\sum{p^3q^3}+30pqr\sum{p^3}+28pqr\sum{p^2(q+r)}-255p^2q^2r^2\geq0$$ which is schur and muirhead.

I did not understood. So tried to break $$\frac{2+p^2q^2r^2-(p^2q^2+q^2r^2+r^2p^2)}{1-p^2q^2r^2}\geq\frac{49}{16}$$Which finally comes to $$65p^2q^2r^2 -16\left(p^2q^2+q^2r^2+r^2p^2\right)-17\geq 0$$ Here i am stuck

Answer 1

By you work we need to prove that: $$(\alpha^2-\alpha+1)p^2q^2r^2+\alpha(p^2q^2+q^2r^2+r^2p^2)\leq(1+\alpha)^2$$ for positives $p$, $q$ and $r$ such that $p+q+r=3$.

Now, let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$.

Thus, we need to prove that $f(w^3)\leq0$, where $f$ is a convex function.

But the convex function gets a maximal value for an extreme value of $w^3$, which by $uvw$ happens in the following cases.

1. $w^3\rightarrow0^+$.

Let $r\rightarrow0^+$.

Thus, we need to prove that: $$\alpha p^2q^2\leq(1+\alpha)^2.$$ Indeed, by AM-GM $$\alpha p^2q^2\leq\alpha\left(\frac{p+q}{2}\right)^4=\frac{81\alpha}{16}$$ and it's enough to prove that $$\frac{81\alpha}{16}\leq(1+\alpha)^2$$ or $$\alpha^2-\frac{49}{16}\alpha+1\geq0,$$ which gives $$\alpha\in\left(-\infty,\frac{49-9\sqrt{17}}{32}\right]\cup\left[\frac{49+9\sqrt{17}}{32},+\infty\right),$$ which is true for $$\alpha\geq \frac{49+9\sqrt{17}}{32}.$$ 2. Two variables are equal.

Let $q=p$ and $r=3-2p$, where $0<p<1.5$.

Thus, we need to prove that: $$(\alpha^2-\alpha+1)p^4(3-2p)^2+\alpha(p^4+2p^2(3-2p)^2)\leq(1+\alpha)^2$$ or $$(p-1)^2(1+2(1+\alpha)^2p+3(\alpha^2-4\alpha+1)p^2+4(\alpha^2-\alpha+1)p^3-4(\alpha^2-\alpha+1)p^4)\geq0,$$ which is true for any $0<p<1.5$ and $\alpha\geq\frac{49+9\sqrt{17}}{32}.$