How to prove $2\left(\sqrt{ab-1}+\sqrt{bc-1}+\sqrt{ca-1}\right)\le \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\sqrt{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}.$?

Question

Question. Let $a,b,c>0: abc=a+b+c+2.$ Prove that$$2\left(\sqrt{ab-1}+\sqrt{bc-1}+\sqrt{ca-1}\right)\le \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\sqrt{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}.$$I am looking for a nice proof by hand, for which there is a possibility to find this proof during a competition.

Equality holds at $a=b=c=2.$

I've tried to use the well-known substitution without success.

Indeed, by the given condition we easily get $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1.$$ Now, if we set $a=\dfrac{y+z}{x};b=\dfrac{x+z}{y}$ then $c=\dfrac{y+x}{z}.$ Replace the set to OP, it's very complicated.

Hope to see some ideas. All contributions are welcome.

Answer 1

I hope the follwing will help.

By squaring both side, we'll prove that $$4(ab+bc+ca-3)+8\sum_{cyc}\sqrt{(ab-1)(bc-1)}\le \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right) \tag{*}.$$ Now, by Schur of third degree inequality $$\sqrt{a}+\sqrt{b}+\sqrt{c}+\frac{9\sqrt{abc}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\ge 2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right),$$or $$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge 4\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)-\frac{9\sqrt{abc}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}>0.$$ Thus, we prove stronger inequality of $(*)$ $$4(ab+bc+ca-3)+8\sum_{cyc}\sqrt{(ab-1)(bc-1)}\le 4\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)^2-\frac{9\sqrt{abc}\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{\sqrt{a}+\sqrt{b}+\sqrt{c}},$$ or $$\frac{9\sqrt{abc}\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+8\sum_{cyc}\sqrt{(ab-1)(bc-1)}\le 8\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+12 \tag{**}.$$

Now, use your subtitution $a=\dfrac{y+x}{z};b=\dfrac{y+z}{x}$ and $c=\dfrac{z+x}{y}$ where $x,y,z>0.$ It implies $ab-1=\dfrac{y(x+y+z)}{xz};bc-1=\dfrac{z(x+y+z)}{xy};ca-1=\dfrac{x(x+y+z)}{yz}.$

Hence, by AM-GM $$8\sum_{cyc}\sqrt{(ab-1)(bc-1)}=8\sum_{cyc}\frac{(x+y+z)}{\sqrt{xy}}\le 8\sum_{cyc}\frac{(x+y)}{\sqrt{xy}}+4\sum_{cyc}\frac{x+y}{z}.\tag{1}$$ Also, by C-S \begin{align*} 8\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)&=8\sum_{cyc}\frac{(y+x)\sqrt{(z+y)(x+z)}}{z\sqrt{yx}}\\&\ge 8\sum_{cyc}\frac{(y+x)(z+\sqrt{yx})}{z\sqrt{yx}}\\&=8\sum_{cyc}\frac{x+y}{\sqrt{xy}}+8\sum_{cyc}\frac{x+y}{z}. \tag{2} \end{align*} From $(1)$ and $(2)$ it remains to prove $$4\sum_{cyc}\frac{x+y}{z}+12\ge \frac{9\sqrt{abc}\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$$ Notice that by C-S \begin{align*} 4\sum_{cyc}\frac{x+y}{z}+12&=2\left[(x+y)+(y+z)+(z+x)\right]\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \\& \ge 2\left(\sum_{cyc}\sqrt{\frac{x+y}{z}}\right)^2\\&=2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2. \end{align*} Thus, we need to prove $$2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^3\ge 9\sqrt{abc}\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right).$$ Since $a+b+c=2\sum_{cyc}\dfrac{x+y}{z}\ge 6$ then $abc=a+b+c+2\le \dfrac{4(a+b+c)}{3}.$

Finally, we prove $$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^3\ge 3\sqrt{3(a+b+c)}\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right),$$which is true by AM-GM \begin{align*} \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^3&=\sqrt{\left[a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\right]^3}\\&\ge \sqrt{27(a+b+c)\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)^2}\\&= 3\sqrt{3(a+b+c)}\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right). \end{align*} Hence, the proof is done.

Answer 2

Not sure where this will lead but let's give it a go!

Let $$A^2 = \frac{1}{a+1}, \quad B^2 = \frac{1}{b+1}, \quad C^2 = \frac{1}{c+1}$$ So that $A^2 + B^2 + C^2 = 1$. (like OP mentioned)

Then we simply get $$a = \frac{1-A^2}{A^2}, \quad b = \frac{1-B^2}{B^2}, \quad c = \frac{1-C^2}{C^2}$$ so $$ab = \left(\frac{1}{A^2}-1\right)\left(\frac{1}{B^2}-1\right)$$ $$ = \frac{1}{A^2B^2}-\frac{1}{A^2}-\frac{1}{B^2}+1$$ $$\sqrt{ab-1} = \sqrt{\frac{1-(A^2+B^2)}{A^2B^2}} = \frac{\sqrt{1-(A^2+B^2)}}{AB}$$

Plugging all of this in and squaring both sides we get $$4\left(\frac{\sqrt{1-(A^2+B^2)}}{AB}+\frac{\sqrt{1-(B^2+C^2)}}{BC}+\frac{\sqrt{1-(A^2+C^2)}}{AC}\right)^2 \leq \,\,\,...$$

$$4\left(\frac{C}{AB}+\frac{A}{BC}+\frac{B}{AC}\right)^2 \leq \left(\frac{\sqrt{1-A^2}}{A}+\frac{\sqrt{1-B^2}}{B}+\frac{\sqrt{1-C^2}}{C}\right)^2\left(\frac{\sqrt{C^2+A^2B^2}}{AB}+\frac{\sqrt{A^2+B^2C^2}}{BC}+\frac{\sqrt{B^2+A^2C^2}}{AC}\right)$$ We can add the terms in the brackets and after simplifying get $$\frac{4}{(ABC)^2} \leq \frac{\left(BC\sqrt{1-A^2}+AC\sqrt{1-B^2}+AB\sqrt{1-C^2}\right)^2}{(ABC)^2}\left(\frac{C\sqrt{C^2+A^2B^2}+B\sqrt{B^2+A^2C^2}+A\sqrt{A^2+B^2C^2}}{ABC}\right)$$ simplifying again $$4ABC \leq \left(BC\sqrt{1-A^2}+AC\sqrt{1-B^2}+AB\sqrt{1-C^2}\right)^2\left(C\sqrt{C^2+A^2B^2}+B\sqrt{B^2+A^2C^2}+A\sqrt{A^2+B^2C^2}\right)$$

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I wasn't sure what the most efficient way from here on out was. I had made a few attempts but they didn't seem to simplify anything any further, so the following is just something that I think might help anyone else also trying to solve this.

I don't think there is much point in showing attempts, and thought it would be better to show everyone how to get back to an inequality in $a, b, c$ alone while I keep attempting at the final solution.

To get it back into a form in terms of $a,b,c$, we can divide both sides by $(ABC)^3$ to obtain

$$\frac{4}{(ABC)^2} \leq \left(\frac{\sqrt{1-A^2}}{A}+\frac{\sqrt{1-B^2}}{B}+\frac{\sqrt{1-B^2}}{B}\right)^2\left(\sqrt{\frac{C^2}{A^2B^2}+1}+\sqrt{\frac{B^2}{A^2C^2}+1}+\sqrt{\frac{A^2}{B^2C^2}+1}\right)$$

Now we can re-write everything in terms of $a, b, c$ again using the identities, and we arrive at

$$4(a+1)(b+1)(c+1) \leq \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)$$

We can simplify the RHS slightly to get

$$4abc+12 \leq \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)$$

Again I would just like to clarify that this is incomplete! If anyone spots a mistake or inconsistency please let me know.

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Edit: The inequality has been solved by a different answer, and I can't seem to get to a solution using this method, so I'll say for now unless I find a way to solve it I probably will not be updating\adding to my approach here anytime soon.

Answer 3

By your work we need to prove that: $$\sum_{cyc}\sqrt{\frac{x+y}{z}}\sqrt{\sum_{cyc}\sqrt{\frac{(x+y)(x+z)}{yz}}}\geq2\sum_{cyc}\sqrt{\frac{(x+y)(x+z)}{yz}-1},$$ where $x$, $y$ and $z$ are positives or $$\sum_{cyc}\sqrt{xy(x+y)}\sqrt{\sum_{cyc}\sqrt{x(x+y)(x+z)}}\geq2\sqrt[4]{xyz}\sqrt{(x+y+z)^3}$$ or $$\sum_{cyc}\left(x^2y+x^2z+2\sqrt{xyz}\sqrt{x(x+y)(x+z)}\right)\sum_{cyc}\sqrt{x(x+y)(x+z)}\geq4\sqrt{xyz}(x+y+z)^3,$$ which gives a quadratic inequality of $\sum\limits_{cyc}\sqrt{x(x+y)(x+z)}$ and we obtain to prove that: $$\sum_{cyc}\sqrt{x(x+y)(x+z)}\geq\frac{\sqrt{\left(\sum\limits_{cyc}(x^2y+x^2z)\right)^2+32 xyz(x+y+z)^3}-\sum\limits_{cyc}(x^2y+x^2z)}{4\sqrt{xyz}}.$$ Now, by C-S we obtain: $$\sum_{cyc}\sqrt{x(x+y)(x+z)}=\sqrt{\sum_{cyc}\left(x(x+y)(x+z)+2(x+y)\sqrt{xy(x+z)(y+z)}\right)}=$$ $$=\sqrt{\sum_{cyc}\left(x(x+y)(x+z)+2(x+y)\sqrt{x^2y^2+xyz(x+y+z)}\right)}=$$ $$=\sqrt{\sum_{cyc}\left(x(x+y)(x+z)+(x+y)\sqrt{(1+3)(x^2y^2+xyz(x+y+z))}\right)}\geq$$ $$=\sqrt{\sum_{cyc}\left(x(x+y)(x+z)+(x+y)xy+(x+y)\sqrt{3xyz(x+y+z)}\right)}=$$ $$=\sqrt{\sum_{cyc}(x^3+2x^2y+2x^2z+xyz)+2\sqrt{3xyz(x+y+z)^3}}$$ and it's enough to prove that: $$\sqrt{\sum_{cyc}(x^3+2x^2y+2x^2z+xyz)+2\sqrt{3xyz(x+y+z)^3}}\geq$$ $$\geq\frac{\sqrt{\left(\sum\limits_{cyc}(x^2y+x^2z)\right)^2+32 xyz(x+y+z)^3}-\sum\limits_{cyc}(x^2y+x^2z)}{4\sqrt{xyz}}.$$ The rest is smooth:

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $f(v^2)\geq0$, where $f$ increases, which by $uvw$ says that it's enough to prove $f(v^2)\geq0$ for equality case of two variables, which gives a right inequality.

Indeed, we need to prove that: $$\sqrt{27u^3-27uv^2+3w^3+18uv^2-6w^3+3w^3+2\sqrt{81u^3w^3}}\geq$$ $$\geq\frac{\sqrt{(9uv^2-3w^3)^2+32\cdot27u^3w^3}-(9uv^2-3w^3)}{4\sqrt{w^3}}$$ or$$\sqrt{3u^3-uv^2+2\sqrt{u^3w^3}}\geq\frac{\sqrt{(3uv^2-w^3)^2+96u^3w^3}-(3uv^2-w^3)}{4\sqrt{w^3}}$$ or $f(v^2)\geq0,$ where $$f(v^2)=\sqrt{3u^3-uv^2+2\sqrt{u^3w^3}}\left(\sqrt{(3uv^2-w^3)^2+96u^3w^3}+3uv^2-w^3\right)-24u^3\sqrt{w^3}.$$ We'll prove that $f(v^2)\geq0$, which is $$-\frac{u\left(\sqrt{(3uv^2-w^3)^2+96u^3w^3}+3uv^2-w^3\right)}{2\sqrt{3u^2-uv^2+2\sqrt{u^3w^3}}}+$$ $$+\sqrt{3u^3-uv^2+2\sqrt{u^3w^3}}\cdot\left(\frac{3u(3uv^2-w^3)}{\sqrt{(3uv^2-w^3)^2+96u^3w^3}}+3u\right)\geq0$$ or $$6\left(3u^3-uv^2+2\sqrt{u^3w^3}\right)\left(\tfrac{3uv^2-w^3}{\sqrt{(3uv^2-w^3)^2+96u^3w^3}}+1\right)\geq$$ $$ \geq\sqrt{(3uv^2-w^3)^2+96u^3w^3}+3uv^2-w^3,$$ for which it's enough to prove that: $$6\left(3u^3-uv^2+2\sqrt{u^3w^3}\right)\geq\sqrt{(3uv^2-w^3)^2+96u^3w^3}+3uv^2-w^3$$ and since by Schur $$3u^3-uv^2\geq3uv^2-w^3,$$ it's enough to prove that $$6(2t+2)\geq\sqrt{4t^2+96}+2t,$$ where $3uv^2-w^3=2t\sqrt{u^3w^3},$ or $$5t+6\geq\sqrt{t^2+24},$$ which is obvious.

Id est, $f$ increases and it's enough to check, what happens for $y=z=1$, which gives $$\sqrt{x^3+4x^2+7x+6+2\sqrt{3x(x+2)^3}}\geq\tfrac{\sqrt{(x^2+x+1)^2+8x(x+2)^3}-x^2-x-1}{2\sqrt{x}}$$ or $$\sqrt{x^3+4x^2+7x+6+2\sqrt{3x(x+2)^3}}\left(\sqrt{(x^2+x+1)^2+8x(x+2)^3}+x^2+x+1\right)\geq4\sqrt{x}(x+2)^3$$ or $$\left(x^2+2x+3+2\sqrt{3x(x+2)}\right)\left(\sqrt{(x^2+x+1)^2+8x(x+2)^3}+x^2+x+1\right)^2\geq16x(x+2)^5$$ or $$2(5x^4+26x^3+51x^2+34x+1)\sqrt{3x(x+2)}+$$ $$+(x^2+x+1)(x^2+2x+3)\sqrt{9x^4+50x^3+99x^2+66x+1}\geq$$$$\geq3x^6+44x^5+202x^4+426x^3+418x^2+152x-3-2(x^2+x+1)\sqrt{3x(x+2)(9x^4+50x^3+99x^2+66x+1)}$$ and since for $$3x^6+44x^5+202x^4+426x^3+418x^2+152x-3-2(x^2+x+1)\sqrt{3x(x+2)(9x^4+50x^3+99x^2+66x+1)}<0$$ the inequality is obvious we can assume that $$3x^6+44x^5+202x^4+426x^3+418x^2+152x-3-2(x^2+x+1)\sqrt{3x(x+2)(9x^4+50x^3+99x^2+66x+1)}\geq0$$ and after squaring of the both sides it's enough to prove(there $4x(x+2)^5$ is canceled!): $$2(x^2+x+1)\sqrt{3x(x+2)(9x^4+50x^3+99x^2+66x+1)}\geq10x^5+48x^4+90x^3+83x^2+42x-3,$$ which is obvious for $10x^5+48x^4+90x^3+83x^2+42x-3<0$.

But for $10x^5+48x^4+90x^3+83x^2+42x-3\geq0$ we have $$3x^6+44x^5+202x^4+426x^3+418x^2+152x-3\geq$$ $$\geq3x^6+44x^5+202x^4+426x^3+418x^2+26x-3+3(-10x^5-48x^4-90x^3-83x^2+3)=$$ $$=3x^6+14x^5+58x^4+156x^3+169x^2+26x+6\geq(x^2+x+1)(3x^4+11x^3+44x^2+40x+4)\geq$$ $$\geq2(x^2+x+1)\sqrt{3x(x+2)(9x^4+50x^3+99x^2+66x+1)},$$ which says that it's enough to prove that: $$12(x^2+x+1)^2x(x+2)(9x^4+50x^3+99x^2+66x+1)\geq(10x^5+48x^4+90x^3+83x^2+42x-3)^2$$ for $10x^5+48x^4+90x^3+83x^2+42x-3\geq0$ or $$(x-1)^2(2x^2+4x+3)(4x^6+36x^5+126x^4+212x^3+183x^2+90x-3)\geq0,$$ which is true because $$4x^6+36x^5+126x^4+212x^3+183x^2+90x-3\geq$$ $$\geq4x^6+36x^5+126x^4+212x^3+183x^2+90x-(10x^5+48x^4+90x^3+83x^2+42x)>0$$ and we are done!