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Question Number 181099 by Ari last updated on 21/Nov/22

$$$$$$\mathrm{prove}\mathrm{that}\mathrm{for}\mathrm{every}$$$$\mathrm{positivenumber}\mathrm{p}\mathrm{e}\mathrm{q}\mathrm{wee}$$$$\mathrm{hav}e:$$$$p+q⩾\sqrt{4pq}$$

Answered by Agnibhoo98 last updated on 21/Nov/22

$$\mathbf{When}\mathit{p}\mathbf{and}\mathit{q}\mathbf{are}\mathbf{positive}\mathbf{numbers}$$$${(p-q)}^2⩾0$$$$\mathrm{or}{p}^2-2pq+q^2⩾0$$$$\mathrm{or}{p}^2-2pq+4pq+q^2⩾4pq\left(\mathbf{Adding}4\mathit{p}\mathit{q}\mathbf{both}\mathbf{side}\right)$$$$\mathrm{or}{p}^2+2pq+q^2⩾4pq$$$$\mathrm{or}{(p+q)}^2⩾4pq$$$$\mathrm{or}p+q⩾\sqrt{4pq}\left(\mathbf{Proved}\right)$$

Commented by Ari last updated on 21/Nov/22

$$ThanksMr\int !$$

Commented by Agnibhoo98 last updated on 22/Nov/22

$$\mathbf{Another}\mathbf{way}$$$$\mathbf{According}\mathbf{to}\mathbf{AM}⩾\mathbf{GM}\mathbf{method}:$$$$\frac{p+q}{2}⩾\sqrt{pq}$$$$\mathrm{or}p+q⩾2\sqrt{pq}$$$$\mathrm{or}p+q⩾\sqrt{4pq}\left(\mathbf{Proved}\right)$$

Answered by SEKRET last updated on 22/Nov/22

$${(\sqrt{\mathbf{p}}-\sqrt{\mathbf{q}})}^2⩾0$$$$\mathbf{p}+\mathbf{q}⩾\sqrt{4\mathbf{pq}}$$

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