paragdey01 wrote: If $a, b$ are positive reals and $a+b<2$ Prove that $\frac{1}{1+a^2}+\frac{1}{1+b^2} \le \frac{2}{1+ab}$ .and determine all $a, b$ yielding equality Because $$\frac{2}{1+ab}-\frac{1}{1+a^2}-\frac{1}{1+b^2}=\frac{(a-b)^2(1-ab)}{(1+ab)(1+a^2)(1+b^2)}\geq0.$$

Another easy computation yields. On breaking we get $a^2+b^2–2ab(a^2+b^2)+a^2b^2+a^2b^2+ab(a^2+b^2)\ge 0$ .

paragdey01 wrote: If $a, b$ are positive reals and $a+b<2$ Prove that $\frac{1}{1+a^2}+\frac{1}{1+b^2} \le \frac{2}{1+ab}$ .and determine all $a, b$ yielding equality Old $\frac{1}{1+a^2}+\frac{1}{1+b^2}\le \frac{2}{1+ab}\iff (1-ab)(a-b)^2\ge 0.$ Generalization of Austrian 2018: Let $ a_1,a_2,\cdots,a_n$ $(n\ge 2)$ are positive reals such that $ a_1+a_2+\cdots+a_n\leq n.$ Prove that $$\frac{1}{1+a^2_1}+\frac{1}{1+a^2_2}+\cdots+\frac{1}{1+a^2_n}\le \frac{n}{1+a_1a_2\cdots a_n}.$$

paragdey01 wrote: If $a, b$ are positive reals and $a+b<2$ Prove that $\frac{1}{1+a^2}+\frac{1}{1+b^2} \le \frac{2}{1+ab}$ .and determine all $a, b$ yielding equality Proof of Zhangyanzong and Zhangyunhua:

Let $a, b$ are positive reals . Prove that $$\frac{a}{1+a^2}+\frac{b}{1+b^2} \le \frac{a+b}{1+ab}.$$

After simplifying we get $a^3+b^3\ge ab(a+b)$ Which is true.

sqing wrote: Let $a, b$ are positive reals . Prove that $$\frac{a}{1+a^2}+\frac{b}{1+b^2} \le \frac{a+b}{1+ab}.$$ Perhaps overkill for this problem, but an interesting technique anyway. Let $a=\tan\left(\frac{A}{2}\right)$, $b=\tan\left(\frac{B}{2}\right)$. The inequality becomes $\frac{\sin A+\sin B}{2}\le\frac{\sin\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}$, or after using sum to product formulae, $2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\le\frac{2\sin\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}$ or $\cos^2\left(\frac{A-B}{2}\right)\le1$, which is obvious.

sqing wrote: Let $a, b$ are positive reals . Prove that $$\frac{a}{1+a^2}+\frac{b}{1+b^2} \le \frac{a+b}{1+ab}.$$ We have $\begin{array}{l} \frac{{a + b}}{{1 + ab}} - \left( {\frac{a}{{1 + {a^2}}} + \frac{b}{{1 + {b^2}}}} \right) = \frac{a}{{1 + ab}} - \frac{a}{{1 + {a^2}}} + \frac{b}{{1 + ab}} - \frac{b}{{1 + {b^2}}} = \frac{{{a^2}\left( {a - b} \right)}}{{\left( {1 + ab} \right)\left( {1 + {a^2}} \right)}} + \frac{{{b^2}\left( {b - a} \right)}}{{\left( {1 + ab} \right)\left( {1 + {b^2}} \right)}}\\= \frac{{a - b}}{{1 + ab}}\left( {\frac{{{a^2}}}{{1 + {a^2}}} - \frac{{{b^2}}}{{1 + {b^2}}}} \right) = \frac{{\left( {a + b} \right){{\left( {a - b} \right)}^2}}}{{\left( {1 + ab} \right)\left( {1 + {a^2}} \right)\left( {1 + {b^2}} \right)}} \ge 0 \end{array}$

We have $\frac{1}{1+a^2}+\frac{1}{1+b^2} =2- \frac{a^2}{1+a^2}-\frac{b^2}{1+b^2}$ We need to prove that $ \frac{a^2}{1+a^2}+\frac{b^2}{1+b^2} \geq \frac{2ab}{1+ab}$ By Titu's Lemma $ \frac{a^2}{1+a^2}+\frac{b^2}{1+b^2} \geq \frac{(a+b)^2}{2+a^2+b^2}$ and the inequality we have to prove is equivalent to : $(a+b)^2(1+ab)\geq 2ab(2+a^2+b^2)$ <=>$(a-b)^2(ab-1)\leq 0$ But $2>a+b\geq 2\sqrt ab$.So $ab <1$ and we are done

paragdey01 wrote: If $a, b$ are positive reals and $a+b<2$ Prove that $\frac{1}{1+a^2}+\frac{1}{1+b^2} \le \frac{2}{1+ab}$ .and determine all $a, b$ yielding equality $\frac{1}{1+a^2}+\frac{1}{1+b^2}=1+\frac{1-a^2b^2}{(1+a^2)(1+b^2)} \le 1+\frac{1-a^2b^2}{(1+ab)^2}=\frac{2}{1+ab}$

Austrian Mathematics Olympiad 2016: Let $a, b, c$ and $d$ be real numbers with $ a^2 + b^2 + c^2 + d^2 = 4.$ Prove that the inequality $$(a + 2)(b + 2) \geq cd.$$holds and give four numbers $a, b, c$ and $d$ such that equality holds. (Proposed by Walther Janous) Austrian Mathematics Olympiad 2015: Let $a, b, c$ and $d$ be positive numbers. Prove that$$ (a^2 + b^2 + c^2 + d^2)^2 \geq (a + b)(b + c)(c + d)(d + a).$$When does equality hold? (Proposed by Georg Anegg)

sqing wrote: Austrian Mathematics Olympiad 2016: Let $a, b, c$ and $d$ be real numbers with $ a^2 + b^2 + c^2 + d^2 = 4.$ Prove that the inequality $$(a + 2)(b + 2) \geq cd.$$holds and give four numbers $a, b, c$ and $d$ such that equality holds. (Proposed by Walther Janous) Rewriting, we need to prove $\left(a+2\right)\left(b+2\right)\ge\frac{4-a^2-b^2-\left(c-d\right)^2}{2}$ This is equivalent to $\frac{\left(a+b+2\right)^2+\left(c-d\right)^2}{2}\ge0$

Kaskade wrote: sqing wrote: Austrian Mathematics Olympiad 2016: Let $a, b, c$ and $d$ be real numbers with $ a^2 + b^2 + c^2 + d^2 = 4.$ Prove that the inequality $$(a + 2)(b + 2) \geq cd.$$holds and give four numbers $a, b, c$ and $d$ such that equality holds. (Proposed by Walther Janous) Rewriting, we need to prove $\left(a+2\right)\left(b+2\right)\ge\frac{4-a^2-b^2-\left(c-d\right)^2}{2}$ This is equivalent to $\frac{\left(a+b+2\right)^2+\left(c-d\right)^2}{2}\ge0$ Thanks. here

sqing wrote: Austrian Mathematics Olympiad 2015: Let $a, b, c$ and $d$ be positive numbers. Prove that$$ (a^2 + b^2 + c^2 + d^2)^2 \geq (a + b)(b + c)(c + d)(d + a).$$When does equality hold? (Proposed by Georg Anegg) $$(a^2 + b^2 + c^2 + d^2)^2 \geq 4(a^2 + b^2 )( c^2 + d^2)\geq(a + b)^2(c + d)^2,$$$$(a^2 + b^2 + c^2 + d^2)^2 \geq4(b^2 + c^2 )( d^2 + a^2)\geq(b + c)^2(d + a)^2, ...$$Let $a, b, c$ be positive numbers. Prove that$$ (a^2 + b^2 + c^2 )^{\frac{3}{2}} \geq \frac{3\sqrt{3}}{8}(a + b)(b + c)(c + a).$$(Zhangyanzong)

sqing wrote: Let $a, b$ are positive reals . Prove that $$\frac{a}{1+a^2}+\frac{b}{1+b^2} \le \frac{a+b}{1+ab}.$$ Equivalent to showing $\frac{(ab+1)(a+b)}{(a^2+1)(b^2+1)} \le \frac{a+b}{ab+1}$ Simplifying and clearing reduces to $a^2 + b^2 \ge 2ab$ which is AM-GM!

sqing wrote: Thank you very much. Let $a, b$ are positive reals . Prove that $$\frac{a}{1+a^2}+\frac{b}{1+b^2} \le \frac{a+b}{1+ab}.$$ This inequality can be proved in such an interesting way. When $a, b< 1$ let us use the substitution $a=th(x)$ and $b=th(y)$. Then our inequality transforms to the following one: $$\frac 1 2 th(2x) + \frac 1 2th(2y) \leq th(x+y).$$But this is just a Jensen's inequality for the function $f(t)=th(2t)$. Case $a,b > 1$ is reduced to the previous one after the changing of variables $a_1=1/a$, $b_1=1/b$. Finally, when $a > 1 > b$, let us take $a_1=1/a$. Then $$\frac{a}{1+a^2}+\frac{b}{1+b^2} =\frac{a_1}{1+a_1^2}+\frac{b}{1+b^2} \leq \frac{a_1+b}{1+a_1b} \leq\frac{a+b}{1+ab}.$$The last inequality can be proved by direct computations: $$\frac{a_1+b}{1+a_1b} \leq\frac{a+b}{1+ab} \quad \Leftrightarrow\quad (a-a_1)(1-b^2)\geq 0.$$Case $a=1$ is trivial.

Let $a,b$ be positive reals such that $a+b=2.$ Prove that $$\frac{1}{a^2+b}+\frac{1}{b^2+a}+\frac{1}{1+ab}\le \frac{9}{2(2+ab)}$$ sqing wrote: Let $a, b$ are positive reals . Prove that $$\frac{a}{1+a^2}+\frac{b}{1+b^2} \le \frac{a+b}{1+ab}.$$ $$\frac{a}{1+a^2}+\frac{b}{1+b^2}=\frac{(a+b)(1+ab)}{(1+a^2)(1+b^2)} \le \frac{(a+b)(1+ab)}{(1+ab)^2}= \frac{a+b}{1+ab}.$$(Zhangyanzong)

Wrong solution

Above I think you applied titu lemma wrong direction

from a + b < 2 we have ab < 1. 2 /1+ab - 1 /1+a^2 - 1 /1+b^2 = (a-b)^2(1-ab) /(1+ab)(1+a^2)(1+b^2) so we have to prove (a-b)^2(1-ab) ≥ 0. (a-b)^2 is square so (a-b)^2 ≥ 0. (1-ab) > 0 because we have ab < 1. so we have proved the inequality and it yields for every a and b such a=b and a+b<2

If $a, b$ are positive reals such that $a+b\leq 2$. Prove that $$\frac{1}{2+a^2}+\frac{1}{2+b^2} \le \frac{4}{3(1+ab)}$$$$\frac{1}{2+a^2}+\frac{1}{2+b^2} \le \frac{2}{1+2ab}$$

sqing wrote: If $a, b$ are positive reals such that $a+b\leq 2$. Prove that $$\frac{1}{2+a^2}+\frac{1}{2+b^2} \le \frac{2}{1+2ab}$$

By AM-GM: $$ab\leq (\frac{a+b}{2})^2<1.$$Observe: \begin{align*} \frac{1}{1+a^2}+\frac{1}{1+b^2} \le \frac{2}{1+ab} \\ \iff 1+\frac{1-a^2b^2}{(1+a^2)(1+b^2)} \leq 1+ \frac{1-a^2b^2}{(1+ab)^2} =\frac{2}{1+ab}.\end{align*}However this is true since by CS: $$(1+a^2)(1+b^2)\geq (1+ab)^2.$$Equality when: $1>a=b\geq 0.$ $\blacksquare$

Let $a, b>0 $ and $ab\leq k$ $(k>0)$. Prove that$$\frac{1}{k+a^2}+\frac{1}{k+b^2} \le \frac{2}{k+ab}.$$

sqing wrote: Let $a, b>0 $ and $ab\leq k$ $(k>0)$. Prove that$$\frac{1}{k+a^2}+\frac{1}{k+b^2} \le \frac{2}{k+ab}.$$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$\frac{1}{k+a^2}+\frac{1}{k+b^2} \le \frac{2}{k+ab}$$$$\Leftarrow (2k+a^2+b^2)(k+ab)\le 2(k+a^2)(k+b^2)$$$$\Leftarrow 2k^2+2abk+a^2k+b^2k+(a^2+b^2)(ab)\le 2k^2+2a^2k+2b^2k+2a^2b^2$$$$\Leftarrow 0\le (a^2+b^2)(k-ab)+2ab(ab-k)$$By AM-GM: $$\Leftarrow 0\le 2ab(0)_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$