If $a,b,c\ge0:a+b+c=3,$ prove $\sqrt{a+b+b^2}+\sqrt{b+c+c^2}+\sqrt{c+a+a^2}\ge 3\sqrt{3}.$

Question

Problem. Let $a,b,c\ge0:a+b+c=3.$ Prove that $$\sqrt{a+b+b^2}+\sqrt{b+c+c^2}+\sqrt{c+a+a^2}\ge 3\sqrt{3}.$$ Equality holds at $(1,1,1);(0,0,3).$

I tried to use some classical inequality but it seems hard to apply. The following is just a thought:

By AM-GM$$\sum_{cyc}\sqrt{a+b+b^2}=\frac{1}{\sqrt{3}}\sum_{cyc}\sqrt{3(a+b+b^2)}\ge 2\sqrt{3}\sum_{cyc}\frac{a+b+b^2}{a+b+b^2+3},$$which is not good enough.

I think we can find suitable yields $a+b+b^2=f(a,b,c)$ and save two of equality cases.

Hope to see some ideas. Thanks for your interest.

Answer 1

We need to prove that: $$\sum_{cyc}\sqrt{4a^2+c^2+ab+2ac+bc}\geq3(a+b+c)$$ or after squaring of the both sides $$\sum_{cyc}\sqrt{(4a^2+c^2+ab+2ac+bc)(4b^2+a^2+2ab+ac+bc)}\geq\sum_{cyc}(2a^2+7ab)$$ or after squaring of the both sides $$2\sum_{cyc}\sqrt{(4a^2+c^2+ab+2ac+bc)\prod_{cyc}(4a^2+c^2+ab+2ac+bc)}\geq$$ $$\geq\sum_{cyc}(14a^3b+17a^3c+31a^2b^2+100a^2bc).$$ But by AM-GM $$2\sum_{cyc}\sqrt{(4a^2+c^2+ab+2ac+bc)\prod_{cyc}(4a^2+c^2+ab+2ac+bc)}\geq$$ $$\geq6\sqrt[3]{\prod_{cyc}(4a^2+c^2+ab+2ac+bc)^2}$$ and it's enough to prove that: $$216\prod_{cyc}(4a^2+c^2+ab+2ac+bc)^2\geq\left(\sum_{cyc}(14a^3b+17a^3c+31a^2b^2+100a^2bc)\right)^3,$$ which is true by BW:

https://www.wolframalpha.com/input?i=216%284x%5E2%2Bz%5E2%2B2xz%2Bxy%2Byz%29%5E2%284y%5E2%2Bx%5E2%2B2xy%2Bxz%2Byz%29%5E2%284z%5E2%2By%5E2%2B2yz%2Bxy%2Bxz%29%5E2-%2814%28x%5E3y%2By%5E3z%2Bz%5E3x%29%2B17%28x%5E3z%2By%5E3x%2Bz%5E3y%29%2B31%28x%5E2y%5E2%2Bx%5E2z%5E2%2By%5E2z%5E2%29%2B100xyz%28x%2By%2Bz%29%29%5E3%2Cx%3Da%2Cy%3Da%2Bu%2Cz%3Da%2Bv

Answer 2

Some thoughts.

The desired inequality is written as $$\sum_{\mathrm{cyc}} \sqrt{\frac{a + b + b^2}{3}} \ge 3. \tag{1}$$

We use the isolated fudging.

We can prove that, for all $a, b \ge 0$ with $a + b \le 3$, $$\sqrt{\frac{a + b + b^2}{3}} \ge \frac{6a^{3} + 9{a}^{2}b + 9a{b}^{2} + 3{b}^{3} - 37{a}^{2} - 58ab - 19{ b}^{2} + 117a + 150b }{180}. \tag{2}$$ (This inequality is true which is verified by Mathematica.)

Using (2), we have \begin{align*} &\sum_{\mathrm{cyc}} \sqrt{\frac{a + b + b^2}{3}}\\[6pt] \ge{}& \sum_{\mathrm{cyc}} \frac{6a^{3} + 9{a}^{2}b + 9a{b}^{2} + 3{b}^{3} - 37{a}^{2} - 58ab - 19{ b}^{2} + 117a + 150b }{180}\\[6pt] ={}& 3 + \frac{1}{180}(a + b + c - 3)(9a^2 + 9b^2 + 9c^2 - 29a - 29b - 29c + 180)\\[6pt] ={}& 3. \end{align*}

Answer 3

My second proof.

The desired inequality is written as $$\sum_{\mathrm{cyc}} \sqrt{\frac{a + b + b^2}{3}} \ge 3. \tag{1}$$

We have $\frac{(a + b) + b^2}{3} \le \frac{3 + 3^2}{3} = 4$, $\frac{(b + c) + c^2}{3} \le \frac{3 + 3^2}{3} = 4$, and $\frac{(c + a) + a^2}{3} \le \frac{3 + 3^2}{3} = 4$.

We use the following bound: for all $x\in [0, 4]$, $$\sqrt x \ge \frac14 x + \frac98 - \frac{9}{16x + 8}. \tag{2}$$ (Note: $\mathrm{LHS}^2 - \mathrm{RHS}^2 = \frac{x(4 - x)(x - 1)^2}{4(2x + 1)^2}\ge 0$.)

Using (2), it suffices to prove that $$\sum_{\mathrm{cyc}} \left(\frac14\cdot \frac{a + b + b^2}{3} + \frac98 - \frac{9}{16\cdot \frac{a + b + b^2}{3} + 8}\right) \ge 3. \tag{3}$$

After homogenization, letting $Q = (a + b + c)/3$, it suffices to prove that $$\sum_{\mathrm{cyc}} \left(\frac14\cdot \frac{aQ + bQ + b^2}{3Q^2} + \frac98 - \frac{9}{16\cdot \frac{aQ + bQ + b^2}{3Q^2} + 8}\right) \ge 3 $$ or (after clearing the denominators) \begin{align*} &15\,{a}^{7}b+9\,{a}^{7}c+45\,{a}^{6}{b}^{2}-40\,{a}^{6}bc+156\,{a}^{6} {c}^{2}+156\,{a}^{5}{b}^{3}-215\,{a}^{5}{b}^{2}c\\ &+46\,{a}^{5}b{c}^{2}+ 288\,{a}^{5}{c}^{3}+267\,{a}^{4}{b}^{4}-247\,{a}^{4}{b}^{3}c-104\,{a}^ {4}{b}^{2}{c}^{2}-88\,{a}^{4}b{c}^{3}\\ &+267\,{a}^{4}{c}^{4}+288\,{a}^{3} {b}^{5}-88\,{a}^{3}{b}^{4}c-288\,{a}^{3}{b}^{3}{c}^{2}-288\,{a}^{3}{b} ^{2}{c}^{3}-247\,{a}^{3}b{c}^{4}\\ &+156\,{a}^{3}{c}^{5}+156\,{a}^{2}{b}^{ 6}+46\,{a}^{2}{b}^{5}c-104\,{a}^{2}{b}^{4}{c}^{2}-288\,{a}^{2}{b}^{3}{ c}^{3}-104\,{a}^{2}{b}^{2}{c}^{4}\\ &-215\,{a}^{2}b{c}^{5}+45\,{a}^{2}{c}^ {6}+9\,a{b}^{7}-40\,a{b}^{6}c-215\,a{b}^{5}{c}^{2}-247\,a{b}^{4}{c}^{3 }-88\,a{b}^{3}{c}^{4}\\ &+46\,a{b}^{2}{c}^{5}-40\,ab{c}^{6}+15\,a{c}^{7}+ 15\,{b}^{7}c+45\,{b}^{6}{c}^{2}+156\,{b}^{5}{c}^{3}+267\,{b}^{4}{c}^{4 }\\ &+288\,{b}^{3}{c}^{5}+156\,{b}^{2}{c}^{6}+9\,b{c}^{7}\\ &\ge 0.\tag{4} \end{align*}

The Buffalo Way (BW) kills it. In detail, WLOG, assume that $c = \min(a, b, c)$ and let $b = c + s, a = c + t$ for $s, t \ge 0$. (4) is written as \begin{align*} &\left( 4374\,{s}^{2}-4374\,st+4374\,{t}^{2} \right) {c}^{6}\\ &+ \left( 6723\,{s}^{3}+6318\,{s}^{2}t-243\,s{t}^{2}+6723\,{t}^{3} \right) {c}^{ 5}\\ &+ \left( 4725\,{s}^{4}+12825\,{s}^{3}t+12555\,{s}^{2}{t}^{2}+1890\,s {t}^{3}+4725\,{t}^{4} \right) {c}^{4}\\ &+ \left( 1745\,{s}^{5}+8449\,{s}^ {4}t+16082\,{s}^{3}{t}^{2}+7658\,{s}^{2}{t}^{3}+1726\,s{t}^{4}+1745\,{ t}^{5} \right) {c}^{3}\\ &+ \left( 329\,{s}^{6}+2562\,{s}^{5}t+8124\,{s}^{ 4}{t}^{2}+7084\,{s}^{3}{t}^{3}+2697\,{s}^{2}{t}^{4}+699\,s{t}^{5}+329 \,{t}^{6} \right) {c}^{2}\\ &+ ( 24s^{7}+335s^{6}t+1846s^ {5}t^{2}+2420s^{4}{t}^{3}+1601s^{3}{t}^{4}+523s^{2}{t}^{ 5}+155st^{6}+24t^{7} ) c\\ &+9\,{s}^{7}t+156\,{s}^{6}{t}^{2} +288\,{s}^{5}{t}^{3}+267\,{s}^{4}{t}^{4}+156\,{s}^{3}{t}^{5}+45\,{s}^{ 2}{t}^{6}+15\,s{t}^{7}\\ &\ge 0. \tag{5} \end{align*} It is very easy to prove that all the coefficients of $c^6, c^5, c^4, c^3, c^2, c, c^0$ are non-negative.

We are done.

Answer 4

My third proof (Sketch).

Fact 1: Let $x, y, z \ge 0$ such that $$3(x^2y^2 + y^2z^2 + z^2x^2) + x^2y^2z^2(x^2 + y^2 + z^2) \ge 12, $$ and $$(x^2 + y^2 + z^2)(3 + 2x^2y^2z^2) \ge 15.$$ Then, $x + y + z \ge 3$.

$\phantom{2}$

Let $$x = \sqrt{(a + b + b^2)/3}, \quad y = \sqrt{(b + c + c^2)/3}, \quad z = \sqrt{(c + a + a^2)/3}.$$ We can prove that $$3(x^2y^2 + y^2z^2 + z^2x^2) + x^2y^2z^2(x^2 + y^2 + z^2) \ge 12$$ and $$(x^2 + y^2 + z^2)(3 + 2x^2y^2z^2) \ge 15.$$

By Fact 1, we have $x + y + z \ge 3$.

We are done.