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If-0-lt-a-b-lt-pi-then-prove-that-sin-ab-sin-a-b-2-32a-2-b-2-ab-a-b-5-




Question Number 156448 by MathSh last updated on 11/Oct/21
If  0<a≤b<π  then prove that:  ((sin(√(ab)))/(sin(((a+b)/2)))) ≥ ((32a^2 b^2 (√(ab)))/((a+b)^5 ))
$$\mathrm{If}\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b}<\pi\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{sin}\sqrt{\mathrm{ab}}}{\mathrm{sin}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right)}\:\geqslant\:\frac{\mathrm{32a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \sqrt{\mathrm{ab}}}{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{5}} } \\ $$
Answered by ghimisi last updated on 11/Oct/21
f(t)=((sint)/t^5 ) ↘  ((a+b)/2)≥(√(ab))⇒f(((a+b)/2))≤f((√(ab)))
$${f}\left({t}\right)=\frac{{sint}}{{t}^{\mathrm{5}} }\:\searrow \\ $$$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}}\Rightarrow{f}\left(\frac{{a}+{b}}{\mathrm{2}}\right)\leqslant{f}\left(\sqrt{{ab}}\right) \\ $$
Commented by MathSh last updated on 11/Oct/21
Very nice dear Ser, thankyou
$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thankyou} \\ $$

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