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Question Number 191301 by Mingma last updated on 22/Apr/23

Answered by a.lgnaoui last updated on 23/Apr/23

$$\mathrm{a};\mathrm{b};\mathrm{c}>0\Rightarrow (\mathrm{a}+\mathrm{b}+\mathrm{c}>0\mathrm{et}\mathrm{abc}>0)$$$$\mathrm{poxons}\mathrm{x}=\mathrm{abc}\mathrm{y}=\mathrm{a}+\mathrm{b}+\mathrm{c}\mathrm{xy}=1$$$$\mathbf{a}+\mathbf{b}=\mathbf{y}-\mathbf{c}\mathbf{a}+\mathbf{c}=\mathbf{y}-\mathbf{b}$$$$(\mathbf{y}-\mathbf{c})(\mathbf{y}-\mathbf{b})={\mathbf{y}}^2-(\mathbf{b}+\mathbf{c})\mathbf{y}+\mathbf{bc}$$$$\Rightarrow \mathbf{ay}+\frac{\mathbf{x}}{\mathbf{a}}\mathbf{xy}=1\Rightarrow \mathbf{x}=\frac{1}{\mathbf{y}}$$$$\mathbf{ay}+\frac{1}{\mathbf{ay}}$$$$\mathbf{ay}=\mathbf{z}(\mathbf{a}+\mathbf{c})(\mathbf{a}+\mathbf{b})=\mathbf{z}+\frac{1}{\mathbf{z}}$$$$\mathbf{Extremum}\mathbf{de}\mathbf{P}\left(\mathbf{z}\right)\mathbf{est}:$$$$\frac{\mathbf{d}(\mathbf{P}\left(\mathbf{z}\right)}{\mathbf{dz}}=01-\frac{1}{{\mathbf{z}}^2}=0\mathrm{z}>0\Rightarrow \mathbf{z}=1$$$$\mathbf{ay}=1\mathbf{y}=\frac{1}{\mathbf{a}}\mathbf{P}\left(\mathbf{z}\right)=2$$$$\mathbf{Donc}\mathbf{minimum}\mathbf{de}(\mathbf{a}+\mathbf{b})(\mathbf{a}+\mathbf{c})\mathbf{est}:2$$

Answered by mr W last updated on 23/Apr/23

$$(a+b)(a+c)$$$$=(a+b+c-c)(a+b+c-b)$$$$=(\frac{1}{abc}-c)(\frac{1}{abc}-b)$$$$={\left(\frac{1}{abc}\right)}^2-(b+c)\frac{1}{abc}+bc$$$$={\left(\frac{1}{abc}\right)}^2-(a+b+c-a)\frac{1}{abc}+bc$$$$={\left(\frac{1}{abc}\right)}^2-(\frac{1}{abc}-a)\frac{1}{abc}+bc$$$$=\frac{1}{bc}+bc$$$$⩾2\sqrt{\frac{1}{bc}\times bc}=2$$$$\Rightarrow minimum=2$$

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