Let $ a=cotA, b=cotB, c=cotC,a'=cotA', b'=cotB', c'=cotC' $ Then $ a,b,c,a',b',c'>0 $ satisfy $ ab+bc+ca=a'b'+b'c'+c'a'=1 $ Suppose the statement isn't true! Thus, we get three inequalities: $ a'(b+c)<\frac{2}{3},b'(a+c)<\frac{2}{3},c'(a+b)<\frac{2}{3} $ Now, $ 1=\sum{a'b'}<\sum{\frac{2}{3(b+c)}\frac{2}{3(c+a)}}=\frac{4}{9}\sum{\frac{1}{(b+c)(c+a)}} $ $ \rightarrow 9(a+b)(b+c)(c+a)<8(ab+bc+ca)(a+b+c) $ , which is impossible for $ a,b,c>0 $.

BSJL wrote: Let $ a=cotA, b=cotB, c=cotC,a'=cotA', b'=cotB', c'=cotC' $ Then $ a,b,c,a',b',c'>0 $ satisfy $ ab+bc+ca=a'b'+b'c'+c'a'=1 $ Suppose the statement isn't true! Thus, we get three inequalities: $ a'(b+c)<\frac{2}{3},b'(a+c)<\frac{2}{3},c'(a+b)<\frac{2}{3} $ Now, $ 1=\sum{a'b'}<\sum{\frac{2}{3(b+c)}\frac{2}{3(c+a)}}=\frac{4}{9}\sum{\frac{1}{(b+c)(c+a)}} $ $ \rightarrow 9(a+b)(b+c)(c+a)<8(ab+bc+ca)(a+b+c) $ , which is impossible for $ a,b,c>0 $. Veery nice. Thank BSJL. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=593684&start=60: $9(a+b)(b+c)(c+a)\ge 8(ab+bc+ca)(a+b+c) $

Let $\triangle ABC $ and $\triangle A'B'C'$ are acute triangles.Prove that\[cotA'(cotB+cotC)+cotB'(cotC+cotA)+cotC'(cotA+cotB)\ge 2.\]

Set $ A'B'C'=XYZ $. We know that $ ctgA= \frac{b^2+c^2-a^2}{4S_{ABC}} $ , where $ S_{ABC} $ is area of the triangle $ ABC $ .similarly ..., we have that $ LHS = \frac{c^2(x^2+y^2-z^2) +a^2(-x^2+y^2+z^2) + b^2(x^2-y^2+z^2)}{ 8S_{ABC} \cdot S_{XYZ} } $ , then we must prove that $ c^2(x^2+y^2-z^2) +a^2(-x^2+y^2+z^2) + b^2(x^2-y^2+z^2) \ge 16S_{ABC} \cdot S_{XYZ} $ , but it's Neuberg-pedoe 's inequality.

Set $ A'B'C'=XYZ $. We know that $ ctgA= \frac{b^2+c^2-a^2}{4S_{ABC}} $ , where $ S_{ABC} $ is area of the triangle $ ABC $ .similarly ..., we have that $ LHS = \frac{c^2(x^2+y^2-z^2) +a^2(-x^2+y^2+z^2) + b^2(x^2-y^2+z^2)}{ 8S_{ABC} \cdot S_{XYZ} } $ , then we must prove that $ c^2(x^2+y^2-z^2) +a^2(-x^2+y^2+z^2) + b^2(x^2-y^2+z^2) \ge 16S_{ABC} \cdot S_{XYZ} $ , but it's Neuberg-pedoe 's inequality.

sqing wrote: Let $\triangle ABC $ and $\triangle A'B'C'$ are acute triangles.Prove that\[cotA'(cotB+cotC)+cotB'(cotC+cotA)+cotC'(cotA+cotB)\ge 2.\] $\sum cotA' \sum cotA=\sqrt{(\sum cot^2A'+2)(\sum cot^2A+2)}\ge\sum cotA'cotA +2.$ $\Rightarrow \sum cotA'(cotB+cotC)\ge 2.$Involves two triangle inequality (6) Pedoe's Inequality Proof of Inequality related to 2 Triangles Like Weitzenboeck inequality but for two triangles

sqing wrote: Let $\triangle ABC $ and $\triangle A'B'C'$ are acute triangles.Prove that\[cotA'(cotB+cotC)+cotB'(cotC+cotA)+cotC'(cotA+cotB)\ge 2.\] $$\Longrightarrow $$ http://www.ssmrmh.ro/wp-content/uploads/2017/07/PP26068.pdf

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sqing wrote: Let $\triangle ABC $ and $\triangle A'B'C'$ are acute triangles.Prove that\[Max\{cotA'(cotB+cotC),cotB'(cotC+cotA),cotC'(cotA+cotB)\}\ge \frac{2}{3}.\] Just using the following known result: $$(y+z)a+(z+x)b+(x+y)c\ge2\sqrt{(xy+yz+zx)(ab+bc+ca)}$$for $(x,y,z)=(\cot A,\cot B,\cot C)$ and $(a,b,c)=(\cot A',\cot B',\cot C')$ and note that $$\cot A\cot B+\cot B\cot C+\cot C\cot A=\cot A'\cot B'+\cot B'\cot C'+\cot C'\cot A'=1$$we obtain immediately the desired result.

sqing wrote: Let $\triangle ABC $ and $\triangle A'B'C'$ are acute triangles.Prove that\[cotA'(cotB+cotC)+cotB'(cotC+cotA)+cotC'(cotA+cotB)\ge 2.\] Usingusing the inequality of VN Murty: $$(y+z)a+(z+x)b+(x+y)c\ge2\sqrt{(xy+yz+zx)(ab+bc+ca)}$$