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Question Number 72434 by aliesam last updated on 28/Oct/19

Answered by mind is power last updated on 28/Oct/19

$$\frac{\mathrm{c}}{\sin \left(\mathrm{x}\right)}=\frac{\mathrm{d}}{\sin \left(\mathrm{y}\right)}=\frac{\mathrm{BD}}{\sin \left(40\right)}$$$${\mathrm{BD}}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{abcos}\left(40\right)$$$$\frac{\sin \left(\mathrm{y}\right)\mathrm{BD}}{\sin \left(40\right)}=\mathrm{d},\mathrm{c}=\frac{\sin \left(\mathrm{x}\right).\mathrm{bD}}{\sin \left(40\right)}$$$${(\mathrm{a}+\mathrm{b})}^2⩾3(\mathrm{ab}+\mathrm{cd})\iff {\mathrm{a}}^2+{\mathrm{b}}^2-\mathrm{ab}⩾3\mathrm{cd}$$$$({\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{abcos}\left(40\right))={\mathrm{BD}}^2⩽{\mathrm{a}}^2+{\mathrm{b}}^2-\mathrm{ab}$$$$2\cos \left(40\right)⩾1$$$$\mathrm{cd}=\frac{{\mathrm{BD}}^2{\sin }^{\mathrm{z}}\left(\mathrm{x}\right){\sin }^2\left(\mathrm{y}\right)}{{\sin }^2\left(40\right)}$$$$3\mathrm{cd}=\frac{{3\mathrm{BD}}^2{\sin }^2\left(\mathrm{x}\right){\sin }^2\left(\mathrm{y}\right)}{{\sin }^2\left(40\right)}⩽{\mathrm{BD}}^2$$$$\iff {3\sin }^2\left(\mathrm{x}\right){\sin }^2\left(\mathrm{y}\right)⩽{\sin }^2\left(40\right)$$$$\mathrm{x}+\mathrm{y}=40$$$$\Rightarrow \mathrm{x}\mathrm{or}\mathrm{y}⩽20$$$$\sin \left(\mathrm{x}\right)⩽\sin \left(20\right)⩽\sin \left(30\right)=\frac{1}{2}⩽\frac{1}{\sqrt{3}}$$$$\sin \left(\mathrm{y}\right)⩽\sin \left(40\right)$$$$$$$$\Rightarrow {3\sin }^2\left(\mathrm{x}\right){\sin }^2\left(\mathrm{y}\right)⩽3.{\left(\frac{1}{\sqrt{3}}\right)}^2.{\sin }^2\left(40\right)={\sin }^2\left(40\right)$$$$\Rightarrow 3\mathrm{cd}⩽{\mathrm{DB}}^2⩽{\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2\iff 3(\mathrm{cd}+\mathrm{ab})⩽{(\mathrm{a}+\mathrm{b})}^2$$$$$$$$$$$$$$

Commented by aliesam last updated on 28/Oct/19

$$godblessyousir.nicesolution$$

Commented by mind is power last updated on 28/Oct/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}$$

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