anyone ?????????

Never bump so fast, please. \[ \sum_{cyc} \frac{a + b}{\sqrt{ab + 1}} \le \sqrt{3 \left( \sum_{cyc} \frac{(a + b)^2}{ab + 1} \right) }\]It suffices to prove that \[ \sum_{cyc} \frac{(a+ b)^2}{ab + 1} \le 3 \]Which is true since \[ \sum_{cyc} \frac{(a + b)^2}{ab + 1} = 6 - \sum_{cyc} \frac{c^2 + 1}{ab + 1} \le 3 \]And using the fact that \[ \sum_{cyc} \frac{c^2 + 1}{ab + 1} \ge 3 \sqrt[3]{\frac{a^2 + 1}{bc + 1} \cdot \frac{b^2 + 1}{ca + 1} \cdot \frac{c^2 + 1}{ab + 1} } \ge 3 \]NB : \[ ((a^2 + 1)(b^2 + 1)(c^2 + 1))^2 = \prod_{cyc} (a^2 + 1)(b^2 + 1) \ge \prod_{cyc} (ab + 1)^2 = ((ab + 1)(bc + 1)(ca + 1))^2 \]

IndoMathXdZ wrote: Never bump so fast, please. \[ \sum_{cyc} \frac{a + b}{\sqrt{ab + 1}} \le \sqrt{3 \left( \sum_{cyc} \frac{(a + b)^2}{ab + 1} \right) }\]It suffices to prove that \[ \sum_{cyc} \frac{(a+ b)^2}{ab + 1} \le 3 \]Which is true since \[ \sum_{cyc} \frac{(a + b)^2}{ab + 1} = 6 - \sum_{cyc} \frac{c^2 + 1}{ab + 1} \le 3 \]And using the fact that \[ \sum_{cyc} \frac{c^2 + 1}{ab + 1} \ge 3 \sqrt[3]{\frac{a^2 + 1}{bc + 1} \cdot \frac{b^2 + 1}{ca + 1} \cdot \frac{c^2 + 1}{ab + 1} } \ge 3 \]NB : \[ ((a^2 + 1)(b^2 + 1)(c^2 + 1))^2 = \prod_{cyc} (a^2 + 1)(b^2 + 1) \ge \prod_{cyc} (ab + 1)^2 = ((ab + 1)(bc + 1)(ca + 1))^2 \] What inequality did you use here?

Arithmetic, quadratic mean is the first one, then arithmetic geometric mean, and finally Cauchy - Schwarz

Related : If $a,b,c$ are positive real numbers with $abc=1$ then show that $$\frac{a}{a+2}+\frac{b}{b+2} + \frac{c}{c+2} \ge 1$$

adityaguharoy wrote: Related : If $a,b,c$ are positive real numbers with $abc=1$ then show that $$\frac{a}{a+2}+\frac{b}{b+2} + \frac{c}{c+2} \ge 1$$ $\frac{a}{a+2}=1-\frac{2}{a+2}$ so $3-(\frac{2}{a+2}+\frac{2}{b+2}+\frac{2}{c+2})\geq 1$ $\frac{1}{a+2}+ \frac{1}{b+2}+ \frac{1}{c+2} \leq 2$ Just multiply everything with $(a+2)(b+2)(c+2)$ and use $a=x/y,b=y/z,c=z/x$ and then it's easy

Pray450 wrote: Let $a,b,c$ positive reals such that $a^2+b^2+c^2=1$. Prove that $$\frac{a+b}{\sqrt{ab+1}}+\frac{b+c}{\sqrt{bc+1}}+\frac{c+a}{\sqrt{ac+1}}\le 3$$

Let a,b and c be positive numbers such that $ a^{2}+b^{2}+c^{2}=1$.Prove that: $$\frac{a}{\sqrt {1+bc}}+\frac{b}{\sqrt{1+ac}}+\frac{c}{\sqrt{1+ab}}\leq\frac{3}{2}$$

Pray450 wrote: Let $a,b,c$ positive reals such that $a^2+b^2+c^2=1$. Prove that $$\frac{a+b}{\sqrt{ab+1}}+\frac{b+c}{\sqrt{bc+1}}+\frac{c+a}{\sqrt{ac+1}}\le 3$$ Also, SOS helps immediately.

arqady wrote: Pray450 wrote: Let $a,b,c$ positive reals such that $a^2+b^2+c^2=1$. Prove that $$\frac{a+b}{\sqrt{ab+1}}+\frac{b+c}{\sqrt{bc+1}}+\frac{c+a}{\sqrt{ac+1}}\le 3$$ Also, SOS helps immediately. what is SOS?

Pray450 wrote: what is SOS? Sum of Square.

ktddtk1977 wrote: Pray450 wrote: what is SOS? Sum of Square. Is it QM?

Pray450 wrote: Let $a,b,c$ positive reals such that $a^2+b^2+c^2=1$. Prove that $$\frac{a+b}{\sqrt{ab+1}}+\frac{b+c}{\sqrt{bc+1}}+\frac{c+a}{\sqrt{ac+1}}\le 3$$ where do you get this such as rare things?