ahmedAbd wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq 2$$Are there any triples $(a,b,c)$, for which the equality holds? This problem was proposed by Konstantinos Metaxas (me ).

ahmedAbd wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq 2$$Are there any triples $(a,b,c)$, for which the equality holds? Proposed by Konstantinos Metaxas. Stronger: $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq \frac{3\sqrt{2}}{2} $$

For this inequality: $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq \frac{3\sqrt{2}}{2} $$ First of all: $\sqrt{\frac{a^3}{bc+1}} = \frac{a^2}{\sqrt{a+1}}$, since $abc=1$ So, by Cauchy-Scharz we have: $LHS \geq \frac{(a+b+c)^2}{\sum \sqrt{a+1}}$ It suffices to show that $\frac{(a+b+c)^2}{\sum \sqrt{a+1}} \geq \frac{3 \sqrt{2}}{2}$, which becomes: $4(a+b+c)^2 \geq 3 \sum 2 \sqrt{2(a+1)}$. However, by AM-GM we have that $2 \sqrt{2(a+1)} \leq 2 + a+ 1 = a + 3$. Consequently, we have to show that: $$4(a+b+c)^2 \geq 3(a+b+c +9) = 3(a+b+c) + 27$$ But we have that: $(a+b+c)^2 \geq 3(a+b+c)$ (it boils down to $a+b+c \geq 3$, obvious by AM-GM) and $3(a+b+c)^2 \geq 27 \Leftrightarrow a+b+c \geq 3$, which is obviously true. Adding these two we are done.

spiros_gal wrote: For this inequality: $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq \frac{3\sqrt{2}}{2} $$ First of all: $\sqrt{\frac{a^3}{bc+1}} = \frac{a^2}{\sqrt{a+1}}$, since $abc=1$ So, by Cauchy-Scharz we have: $LHS \geq \frac{(a+b+c)^2}{\sum \sqrt{a+1}}$ It suffices to show that $\frac{(a+b+c)^2}{\sum \sqrt{a+1}} \geq \frac{3 \sqrt{2}}{2}$, which becomes: $4(a+b+c)^2 \geq 3 \sum 2 \sqrt{2(a+1)}$. However, by AM-GM we have that $2 \sqrt{2(a+1)} \leq 2 + a+ 1 = a + 3$. Consequently, we have to show that: $$4(a+b+c)^2 \geq 3(a+b+c +9) = 3(a+b+c) + 27$$ But we have that: $(a+b+c)^2 \geq 3(a+b+c)$ (it boils down to $a+b+c \geq 3$, obvious by AM-GM) and $3(a+b+c)^2 \geq 27 \Leftrightarrow a+b+c \geq 3$, which is obviously true. Adding these two we are done. good solution!

ahmedAbd wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq 2$$Are there any triples $(a,b,c)$, for which the equality holds? Proposed by Konstantinos Metaxas. The problem is equivalent to Mathematical Reflections 5 (2017) S421.

VadimM wrote: ahmedAbd wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq 2$$Are there any triples $(a,b,c)$, for which the equality holds? Proposed by Konstantinos Metaxas. Stronger: $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq \frac{3\sqrt{2}}{2} $$ Using Tangent Line Method: \[ \sum \sqrt{\frac{a^3}{1+bc}}=\sum \sqrt{\frac{a^3}{1+\frac{1}{a}}} \geq \sum \frac{7a-3}{4\sqrt{2}} \geq \frac{3\sqrt{2}}{2}\]

bel.jad5 wrote: VadimM wrote: ahmedAbd wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq 2$$Are there any triples $(a,b,c)$, for which the equality holds? Proposed by Konstantinos Metaxas. Stronger: $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq \frac{3\sqrt{2}}{2} $$ Using Tangent Line Method: \[ \sum \sqrt{\frac{a^3}{1+bc}}=\sum \sqrt{\frac{a^3}{1+\frac{1}{a}}} \geq \sum \frac{7a-3}{4\sqrt{2}} \geq \frac{3\sqrt{2}}{2}\] Or holder

Yes of course! $$\left(\sum_{cyc}\sqrt{\frac{a^3}{1+bc}}\right)^2\sum_{cyc}(1+bc)\geq(a+b+c)^3.$$Thus, it remains to prove that $$(a+b+c)^3\geq4\sum_{cyc}(1+ab),$$which is obvious.

ahmedAbd wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\sqrt{\frac{a^3}{1+bc}}+\sqrt{\frac{b^3}{1+ac}}+\sqrt{\frac{c^3}{1+ab}}\geq 2$$Are there any triples $(a,b,c)$, for which the equality holds? Proposed by Konstantinos Metaxas. Another day another dollar Let $a=e^x, b=e^y, c=e^z$ hence we get $x+y+z=0$ by $abc=1$ Now rewrite the ineq using $abc=1$ as $$\sum_{\text{cyc}} \frac{a^2}{\sqrt{a+1}} \ge 2$$Now lets analyze the function: $$f(x)=\frac{e^{2x}}{\sqrt{e^x+1}}$$Its second derivate is $$f''(x)=\frac{e^{2x}(22e^x+9e^{2x}+16)}{4(e^x+1)^{\frac{5}{2}}}>0 \implies f \; \text{convex on} \; \mathbb R$$Hence Using Jensen's Ineq we get that $$f(x)+f(y)+f(z) \ge 3f \left(\frac{x+y+z}{3} \right) \implies \sum_{\text{cyc}} \frac{e^{2x}}{\sqrt{e^x+1}} \ge \frac{3\sqrt{2}}{2}>2$$Hence the ineq holds and there is no equality case, thus we are done

What is IMEO? Note that the function $f(x)=\sqrt{\frac{x^4}{x+1}}$ is convex. By Jensen, it suffices to prove that \[\frac{x^4}{x+1}\geq\frac{4}{9}\text{ where }3x=a+b+c.\]which is true since $x\geq abc=1\implies 9x^4\geq 4x+4\implies \frac{x^4}{x+1}\geq\frac{4}{9}.$

puntre wrote:

IMEO https://artofproblemsolving.com/community/c6h1492618p9015177

Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\sqrt{\frac{a^3}{1+b+c}}+\sqrt{\frac{b^3}{1+c+a}}+\sqrt{\frac{c^3}{1+a+b}} \geq \sqrt 3$$

sqing wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\sqrt{\frac{a^3}{1+b+c}}+\sqrt{\frac{b^3}{1+c+a}}+\sqrt{\frac{c^3}{1+a+b}} \geq \sqrt 3$$ The following inequality a bit of stronger. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\sqrt{\frac{a}{1+b+c}}+\sqrt{\frac{b}{1+c+a}}+\sqrt{\frac{c}{1+a+b}}\geq \sqrt3.$$

here

Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove or disprove $$\sqrt{\frac{a}{1+a+b}} +\sqrt{\frac{b}{1+b+c}}+\sqrt{\frac{c}{1+c+a}} \geq \sqrt3 $$

arqady wrote: Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\sqrt{\frac{a}{1+b+c}}+\sqrt{\frac{b}{1+c+a}}+\sqrt{\frac{c}{1+a+b}}\geq \sqrt3.$$ The following inequality also has a nice proof. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\sqrt{\frac{a}{1+3b+3c}}+\sqrt{\frac{b}{1+3c+3a}}+\sqrt{\frac{c}{1+3a+3b}}\geq\frac{3}{\sqrt7}.$$

sqing wrote: Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove or disprove $$\sqrt{\frac{a}{1+a+b}} +\sqrt{\frac{b}{1+b+c}}+\sqrt{\frac{c}{1+c+a}} \geq \sqrt3 $$ It's wrong for $a=b$ and $c\rightarrow0^+$.

Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that $$\frac{a}{1+a+b}+\frac{b}{1+b+c}+\frac{c}{1+c+a} \geq 1 $$

sqing wrote:

Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that $$\frac{a}{1+a+b}+\frac{b}{1+b+c}+\frac{c}{1+c+a} \geq 1 $$ Are you sure?

sqing wrote: Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that $$\frac{a}{1+a+b}+\frac{b}{1+b+c}+\frac{c}{1+c+a} \geq 1 $$ $\iff a^2b+b^2c+c^2a+abc\ge 1+a+b+c$ $\iff \frac{a}{c}+\frac{c}{b}+\frac{b}{a}\ge a+b+c $ By AM-GM: $\frac{a}{c}+\frac{a}{c}+\frac{c}{b}\ge 3\sqrt[3]{\frac{a^2}{bc}}=3a$

sqing wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\sqrt{\frac{a^3}{1+b+c}}+\sqrt{\frac{b^3}{1+c+a}}+\sqrt{\frac{c^3}{1+a+b}} \geq \sqrt 3$$

\[\sum{\sqrt{\frac{a^3}{bc+1}}}=\sum{\frac{a^2}{\sqrt{a+1}}}>\sum{\frac{a^2}{\frac{a}{2}+1}}=\sum{\frac{2a^2}{a+2}}=\sum{ (2a-4+\frac{8}{a+2})}\overset{?}{\geq} 2\]Because $\frac{a}{2}+1>\sqrt{a+1}\iff a+2>2\sqrt{a+1}\iff a^2+4a+4>4a+4$ which is true. \[2\sum{a}+8\sum{\frac{1}{a+2}}\overset{?}{\geq} 14\iff \sum{a}+4\sum{\frac{1}{a+2}}\overset{?}{\geq} 7\]Let $a+b+c=3u,ab+bc+ca=3v^2,abc=1$. \[LHS=3u+4(\frac{3v^2+12u+12}{6v^2+12u+9})=3u+\frac{4(v^2+4u+4)}{2v^2+4u+3}\overset{?}{\geq} 7\]\[6uv^2+12u^2+9u+4v^2+16u+16\overset{?}{\geq} 14v^2+28u+21\iff 6uv^2+12u^2\overset{?}{\geq} 10v^2+3u+5\]This is equavilent to $v^2(10-6u)\overset{?}{\leq} 12u^2-3u-5$. We see that $RHS>0$ hence suppose $u<\frac{5}{3}$. \[v^2(10-6u)\leq u^2(10-6u)=-6u^3+10u^2\overset{?}{\leq} 12u^2-3u-5\]The last inequality holds because $6u^3+2u^2-3u-5\geq 0$ follows by $u\geq 1$ as desired.$\blacksquare$