Since nobody has posted a solution yet here's a diagram.

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Leartia wrote: Since nobody has posted a solution yet here's a diagram. Prove that if $r>1$ and $t\in(0,\pi),$ then $$(r^2+1)\sqrt2\geq r\sqrt{r^2-2r\cos t+1}+\sqrt{r^2+2r\cos t+1}.$$

Leartia wrote: Assume that in the $\triangle ABC$ there exists a point $D$ on $BC$ and a line $l$ passing through $A$ such that $l$ is tangent to $(ADC)$ and $l$ bisects $BD.$ Prove that $a\sqrt{2}\geq b+c.$ Let $AE \cap (ABC) = X$, we have $AE.EX = BE.CE = ED.EC = EA^2$ so $E$ is the midpoint of $AX$. This means, if $O$ is the circumcenter, then $\angle OEA = 90^\circ$, or that if $M,N$ are midpoints of sides, then $(AMN)$ intersects $BC$, so we must have $\angle BAC \ge 90^\circ$, so $2a^2 \ge 2(b^2 + c^2) \ge (b+c)^2 \implies \sqrt{2}a \ge b+c$

An algebraic approach

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This problem was proposed by @Leartia and me. Solution 1: Let $E$ be the intersection of $l$ with $BC$. Let $t,$ $u$ and $v$ be the lengths of the segments $AE,$ $BE$ and $EC$ respectively. From Stewart's Theorem we have that $\dfrac{t^2}{uv}=\dfrac{\frac{b^2}{u}+\frac{c^2}{v}}{u+v}-1.$ We use Cauchy-Schwarz inequality and get that $\dfrac{b^2}{u}+\dfrac{c^2}{v}\geq \dfrac{(b+c)^2}{u+v}.$ Furthermore since $t$ is tangent to $(ADC)$ we have that $t^2=|ED|*|EC|=|EB|*|EC|=u*v$. Inequality transforms to $1\geq\dfrac{(b+c)^2}{u+v} -1=\dfrac{(b+c)^2}{a} -1$,which is equivalent to $2a^2\geq (b+c)^2$ and the desired result follows. Solution 2: Let $X$ be the reflection of $A$ around $E$. This gives us that $ABXD$ is parallelogram and since $AX$ is tangent to $(ACD)$ we have that $\angle AXB=\angle XAD=\angle ACD=\angle ACB$, implying $ACXB$ is cyclic. Let $s$ and $l$ be the lengths of $BX$ and $CX$, respectively. Since $AE=EX$ and also from the similarities of triangles on the cyclic quadrilateral, we have that $\frac{s}{t}=\frac{b}{v}$ dhe $\frac{l}{t}=\frac{c}{u}$. From Ptolemy's Theorem we have that $2ta=bs+cl$ ose $2 a=\dfrac{bs+cl}{t}.$ From Cauchy-Schwarz inequality Inequality we have that $\dfrac{bs}{t}+\dfrac{cl}{t}=\dfrac{b^2}{v}+\dfrac{c^2}{u}\geq\dfrac{(b+c)^2}{a}.$ So $2a\geq \dfrac{(b+c)^2}{a}$, implying $a\sqrt{2}\geq b+c.$

Leartia wrote: Assume that in the $\triangle ABC$ there exists a point $D$ on $BC$ and a line $l$ passing through $A$ such that $l$ is tangent to $(ADC)$ and $l$ bisects $BD.$ Prove that $a\sqrt{2}\geq b+c.$ Let $E=BC\cup l$ then By Tangent-Secant theorem we have $\angle ACD=\angle DAE$ Hence $|AE|=\sqrt{(|BC|-|BE|)(|BE|)}, |AD|=\frac{|AC|\sqrt{|BE|}}{\sqrt{|BC|-|BE|}}$ Now as By Apollonius Theorem we have $|AB|^2+|AD|^2=2|AE|^2+2|BE|^2\implies \frac{|AC|^2}{\sqrt{2}(|BC|-|BE|)}+\frac{|AB|^2}{\sqrt{2}{|BE|}}=\sqrt{2}|BC|$ Hence as $\sqrt{2}(|BC|-|BE|)+\sqrt{2}|BE|=\sqrt{2}|BC|$ So By Cauchy we have $\sqrt{2}|BC|\geq |AC|+|AB|$

Maths_1729 wrote: Leartia wrote: Assume that in the $\triangle ABC$ there exists a point $D$ on $BC$ and a line $l$ passing through $A$ such that $l$ is tangent to $(ADC)$ and $l$ bisects $BD.$ Prove that $a\sqrt{2}\geq b+c.$ Let $E=BC\cup l$ then By Tangent-Secant theorem we have $\angle ACD=\angle DAE$ Hence $|AE|=\sqrt{(|BC|-|BE|)(|BE|)}, |AD|=\frac{|AC|\sqrt{|BE|}}{\sqrt{|BC|-|BE|}}$ Now as By Apollonius Theorem we have $|AB|^2+|AD|^2=2|AE|^2+|BE|^2\implies \frac{|AC|^2}{\sqrt{2}(|BC|-|BE|)}+\frac{|AB|^2}{\sqrt{2}{|BE|}}=\sqrt{2}|BC|$ Hence as $\sqrt{2}(|BC|-|BE|)+\sqrt{|BE|}=\sqrt{2}|BC|$ So By Cauchy we have $\sqrt{2}|BC|\geq |AC|+|AB|$ wow ! solution