Prove $a^{\sqrt{\log_ba}}+b^{\sqrt{\log_cb}}+c^{\sqrt{\log_ac}}\geqslant a+b+c$ for $a, b, c > 1$

Question

For $$a,b,c > 1$$ show that $$a^{\sqrt{\log_ba}}+b^{\sqrt{\log_cb}}+c^{\sqrt{\log_ac}}\geqslant a+b+c$$

I know how to show this $$a^{{\log_bc}}+b^{{\log_ca}}+c^{{\log_ab}}\geqslant a+b+c$$ $$a^{\log_bc}+b^{\log_ca}=a^{\log_bc}+a^{\log_cb}\geq2\sqrt{a^{\log_bc}a^{\log_cb}}=2\sqrt{a^{\log_bc+\log_cb}}\geq2\sqrt{a^2}=2a$$ Or this one $$(ab)^{\sqrt{\log_ab}}+(bc)^{\sqrt{\log_bc}}+(ca)^{\sqrt{\log_ca}}\geqslant a^2+b^2+c^2$$ $$a^{\sqrt{\log_ab}}=b^{\sqrt{\log_ba}}$$ $$(ab)^{\sqrt{\log_a}}=a^{\sqrt{\log_ab}}b^{\sqrt{\log_ab}}=b^{\sqrt{\log_ba}}b^{\sqrt{\log_ab}}=b^{\sqrt{\log_ab}+\sqrt{\log_ba}}\geq b^2$$ Or even $$a^{{\log_ba}}+b^{{\log_cb}}+c^{{\log_ac}}\geqslant a+b+c$$ But the first one, I don't know how to start. Please give me a starting point.

Answer 1

It suffices to prove that $$a^{\sqrt{\log_b a}} \ge \frac{3a - b}{2}. \tag{1}$$ Summing cyclically on (1), the desired result follows.

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Proof of (1):

We only need to prove the case that $\frac{3a - b}{2} > 1$.

Taking logarithm, it suffices to prove that, for all $1 < b < 3a - 2$, $$\sqrt{\log_b a}\, \ln a \ge \ln \frac{3a - b}{2}$$ or $$\sqrt{\ln a}\, \ln a \ge \sqrt{\ln b}\,\ln \frac{3a - b}{2}$$ or (squaring both sides) $$\ln^3 a \ge \ln b \cdot \ln \frac{3a - b}{2}\cdot \ln \frac{3a - b}{2}.$$

By AM-GM twice, we have \begin{align*} \ln b \cdot \ln \frac{3a - b}{2}\cdot \ln \frac{3a - b}{2} &\le \left(\frac{\ln b + \ln \frac{3a - b}{2} + \ln \frac{3a - b}{2}}{3}\right)^3\\[6pt] &= \frac{1}{27}\ln^3 \left(b \cdot \frac{3a-b}{2 } \cdot \frac{3a-b}{2}\right)\\[6pt] &\le \frac{1}{27}\ln^3 \left(\left(\frac{b + \frac{3a - b}{2} + \frac{3a - b}{2}}{3}\right)^3\right)\\[6pt] &= \frac{1}{27}\ln^3(a^3) \\ &= \ln^3 a. \end{align*}

We are done.

Answer 2

Some hint (hard to do better than the RiverLi's answer) :

Using derivative try to show :

for $x>1,y>1$ we have :

$$x^{\sqrt{\frac{\ln x}{\ln y}}}-1-\left(x-1\right)\sqrt{\frac{x-1}{y-1}}\geq 0$$

Then we need to show :

$$\sum_{cyc}1+\left(x-1\right)\sqrt{\frac{x-1}{y-1}}-x\geq 0$$

Try the substitution : $u^2=x-1,v^2=y-1,w^2=z-1$

Hope you enjoy it .

Ps: I shall add details later .

Adding details :

A bit of algebra (using the subsitution i give) we need to show :

$$\ln^3(x^2+1)\geq \ln^2(x^3/y+1)\ln(y^2+1)$$

We can prove it using :

Let $x>0,y>0,z>0$

Then we have :

$$(\ln((xyz)^{1/3}+1))^3-\ln(x+1)\ln(y+1)\ln(z+1)\ge 0$$

It's straightforward.It's itself show with the log concavity of :

$$\ln(\ln(e^x+1))$$

Answer 3

$\color{green}{\textbf{Useful inequality.}}$

Firstly, let us consider the function $$f(x,y,z)=x^4z+y^4x+z^4y-xyz(x^2+y^2+z^2),\qquad (x>0,\;y>0,\;z>0),\tag 1$$ which has the gradient $$\bar\nabla f(x,y.z) = \begin{pmatrix} y^4+3x^3z-(3x^2+y^2+z^2)yz\\ z^4+3y^3x-(3y^2+z^2+x^2)zx\\ x^4+3z^3y-(3z^2+x^2+y^2)xy \end{pmatrix},$$ Easily to check, that $$(\bar\nabla f((x,y,z))\cdot (x,y,z) = 5f(x,y,z).$$ Therefore, $\,f(x,y,z)=0\,$in the stationary points.

Taking in account that $$\lim_{x\to \infty} f(x,y,z)= +\infty,\quad \lim_{y\to \infty} f(x,y,z)= +\infty,\quad \lim_{z\to \infty} f(x,y,z)= +\infty,$$ this means, that the inequality $$f(x,y,z)\ge 0,\qquad(x>0,\;y>0,\;z>0)\tag2$$ is satisfied on the edges of the domain and, as the result, in the all points of the domain.

$\color{green}{\textbf{Main proof.}}$

Let $\;\ln a=x^2,\quad \ln b = y^2,\quad \ln c=z^2, \quad x>0,\;y>0,\;z>0,$ then

$$\large a^{\sqrt{\log_ba}}=e^{\ln a \sqrt{\frac {\ln a}{\ln b}}} =e^{\frac{x^3}y}.$$

Taking in account $(2),$ one can get: $$a^{\sqrt{\log_ba}}+b^{\sqrt{\log_cb}}+c^{\sqrt{\log_ac}}-(a+b+c) =e^{^{x^3}/y}+e^{^{y^3}/z}+e^{^{z^3}/x} -\left(e^{x^2}+e^{y^2}+e^{z^2}\right)$$ $$=\sum_{k=0}^\infty \dfrac1{k!}\left(\dfrac{x^{3k}}{y^k} +\dfrac{y^{3k}}{z^k}+\dfrac{z^{3k}}{x^k}-x^{2k}-y^{2k}-z^{2k}\right) =\sum_{k=0}^\infty \dfrac1{k!x^ky^kz^k}\, f\left(x^k,y^k,z^k\right) \color{brown}{\mathbf{\ge 0.}}$$

Proved.

Answer 4

Define functions $g(t) = \exp(\exp(t))$, $f(t_1, t_2, t_3) = g(t_1) + g(t_2) + g(t_3)$.

Let $\displaystyle \begin{cases}x_1 = \ln\ln a\\ x_2 = \ln\ln b\\ x_3 = \ln\ln c\end{cases}$, and set $\begin{cases}y_1 = \frac{3x_1 - x_2}{2}\\ y_2 = \frac{3x_2 - x_3}{2}\\ y_3 = \frac{3x_3 - x_1}{2}\end{cases}$, then we can rewrite the inequality as $$ f\left(y_1, y_2, y_3\right)\geq f(x_1, x_2, x_3). \tag{#}$$

To establish (#), note that

1. $(x_1, x_2, x_3)$ is majorized by $(y_1, y_2, y_3)$, since $$ \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} \frac32&-\frac12&0\\ 0&\frac32&-\frac12\\ -\frac12&0&\frac32 \end{bmatrix}^{-1} \begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix} =\begin{bmatrix} \frac9{13}&\frac3{13}&\frac1{13}\\ \frac1{13}&\frac9{13}&\frac3{13}\\ \frac3{13}&\frac1{13}&\frac9{13} \end{bmatrix} \begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix}, $$ and the matrix in the last term is doubly stochastic.

2. $f$ is Schur-convex, since $g$ is convex.

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Remarks

1. An alternative way is to show that $$g(y_1) = g\left(\frac{3x_1 - x_2}{2}\right) \geq \frac{3g(x_1) - g(x_2)}{2}, \tag{##}$$ (and also the other two inequalities for $g(y_2), g(y_3)$), which can be established from the convexity of $g$. Eq. (##) is indeed Eq. (1) in River Li's nice construction.
2. From the convexity of $g$, we can extend (##) to $$g\left((1+p)x_1 - p x_2\right) \geq (1+p)g(x_1) - p\cdot g(x_2), \quad \text{for all $p > 0$},$$ where (##) is the special case of $p = 1/2$. This says that, essentially, we have $$ a^{(\log_ba)^p}+b^{(\log_cb)^p}+c^{(\log_ac)^p}\geqslant a+b+c, \quad \text{for all $p > 0$ and $a, b, c > 1$.} $$