1Try these too ---
1>
In an acute angled triangle, prove that
1 + cos A + cos B + cosC2 cos A cos B cos C ≥ 10
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If x>0 , 0<p<1 then prove that
(1 + x)p ≤ p (x - 1 ) + 2
Prove or disprove that sin AA + sin BB + sin CC ≥ 9 (4 - 3√3 - 3 .25 )2 Î where A,B,C are the angles of a triangle in radian.
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10 Answers
1
1second one.........
consider
g(x)=px +(2-p)
f(x)=(1+x)p
g'(x)=p
f'(x)=p(1+x)p-1
its quite obvious
1>(1+x)p-1 (as p-1 is negative and x is +ve)
multiply p both sides
p>p(1+x)p-1
g'(x)>f(x)...........(for x >0)
g(0)=2-p >=1
f(0)=1
hence
g(x)>=f(x)
hence proved
and one more thing if p >=1 inequality sign is reversed :O
341first inequality can be proved for acute angled triangle.
Dont know proof for general triangle
1Well i am sorry prophet sir both the first and the next are for acute angled triangles only
and xyz ,great solution, exactly what i had done
341First from Jordan's Inequality, \frac{\sin x}{x} \ge \frac{2}{\pi} when x \in \left(0, \frac{\pi}{2} \right]
(this follows from the fact that \frac{\sin x}{x} monotonically decreases in that interval)
Hence \sum \frac{\sin A}{A} \ge \frac{6}{\pi}
It remains to prove that \frac{6}{\pi} \ge \frac{9}{2 \pi} (4- \sqrt[3] {3} - \sqrt [4] {3}) or
\sqrt[3] {3} +\sqrt [4] {3}\ge \frac{8}{3} = \frac{32}{12}
This follows from \sqrt[3] {3} \ge \frac{17}{12}; \sqrt [4] {3}\ge \frac{15}{12} = \frac{5}{4}
as 3 \times 12^3 > 17^3 (No Pain No Gain :D )
and
3 \times 4^4 > 5^4
341Since cosine is a convex function in the given domain, from Jensen's Inequality
\cos A + \cos B + \cos C \le 3 \cos \left(\frac{A+B+C}{3} \right) = \frac{3}{2}
From AM-GM,
\cos A + \cos B + \cos C \ge 3 (\cos A \cos B \cos C)^{\frac{1}{3}}
From the above two inequalities, we have
(\cos A \cos B \cos C) \le \frac{1}{8} \Rightarrow \frac{1}{ \cos A \cos B \cos C} \ge 8
The given expression is
\frac{1 + \cos A + \cos B + \cos C}{ \cos A \cos B \cos C} = \frac{1}{ \cos A \cos B \cos C} + \sum \frac{1}{ \cos A \cos B }
\frac{1}{ \cos A \cos B \cos C} \ge 8 is already proved
From AM-GM, we have
\sum \frac{1}{ \cos A \cos B } \ge 3 \left(\frac{1}{\cos A \cos B \cos C} \right)^\frac{2}{3} \ge 3 \times 8^\frac{2}{3} = 12
Adding the above two we get
\frac{1 + \cos A + \cos B + \cos C}{ \cos A \cos B \cos C} \ge 20
341A more compact way after deducing that \frac{1}{\cos A \cos B \cos C} \ge 8 is:
\frac{1 + \cos A + \cos B + \cos C}{\cos A \cos B \cos C}= \frac{\frac{1}{2} + \frac{1}{2} + \cos A + \cos B + \cos C}{\cos A \cos B \cos C} \\ \\
\ge 5 \times \left(\frac{1}{2} \right)^{\frac{2}{5}} \times \frac{1}{(\cos A \cos B \cos C)^{\frac{4}{5}}} \ge 5 \times \left(\frac{1}{2} \right)^{\frac{2}{5}} \times 8^\frac{4}{5} = 20
341When in doubt, google.
Anyway: http://planetmath.org/encyclopedia/ProofOfJordansInequality.html
1nice solutions prophet sir thanks for introducing new things :)
i had the same solution as in #7 but i proved cosAcosBcosC ≥ 1/8 in another way
\cos A \cos B \cos C=\frac{1}{2} cos A\left(\cos (B+C)+cos(B-C) \right)
put x=cos(B+c) , y=cos(B-C)
-\frac{1}{2}xy-\frac{1}{2}x^2=\frac{1}{2}\left(x+\frac{y}{2} \right)^2 +\frac{y^2}{8} \le \frac{y^2}{8}\le \frac{1}{8}