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

prove that for every positivenumber p e q wee have: p+q≥(√(4pq))

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

Answered by Agnibhoo98 last updated on 21/Nov/22

When p and q are positive numbers (p − q)^2 ≥ 0 or p^2 − 2pq + q^2 ≥ 0 or p^2 − 2pq + 4pq + q^2 ≥ 4pq (Adding 4pq both side) or p^2 + 2pq + q^2 ≥ 4pq or (p + q)^2 ≥ 4pq or p + q ≥ (√(4pq)) (Proved)

$$\boldsymbol{\mathrm{When}}\:\boldsymbol{{p}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{q}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{numbers}} \\ $$$$\left({p}\:−\:{q}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{0} \\ $$$$\mathrm{or}\:{p}^{\mathrm{2}} \:−\:\mathrm{2}{pq}\:+\:{q}^{\mathrm{2}} \:\geqslant\:\mathrm{0} \\ $$$$\mathrm{or}\:{p}^{\mathrm{2}} \:−\:\mathrm{2}{pq}\:+\:\mathrm{4}{pq}\:+\:{q}^{\mathrm{2}} \:\geqslant\:\mathrm{4}{pq}\:\left(\boldsymbol{\mathrm{Adding}}\:\mathrm{4}\boldsymbol{{pq}}\:\boldsymbol{\mathrm{both}}\:\boldsymbol{\mathrm{side}}\right) \\ $$$$\mathrm{or}\:{p}^{\mathrm{2}} \:+\:\mathrm{2}{pq}\:+\:{q}^{\mathrm{2}} \:\geqslant\:\mathrm{4}{pq} \\ $$$$\mathrm{or}\:\left({p}\:+\:{q}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{4}{pq} \\ $$$$\mathrm{or}\:{p}\:+\:{q}\:\geqslant\:\sqrt{\mathrm{4}{pq}}\:\:\:\:\:\left(\boldsymbol{\mathrm{Proved}}\right) \\ $$

Commented by Ari last updated on 21/Nov/22

Thanks Mr∫!

$${Thanks}\:{Mr}\int! \\ $$

Commented by Agnibhoo98 last updated on 22/Nov/22

Another way According to AM ≥ GM method : ((p + q)/2) ≥ (√(pq)) or p + q ≥ 2(√(pq)) or p + q ≥ (√(4pq)) (Proved)

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

Answered by SEKRET last updated on 22/Nov/22

((√(p )) − (√q) )^2 ≥ 0 p+q≥ (√(4pq))

$$\left(\sqrt{\boldsymbol{\mathrm{p}}\:}\:\:−\:\sqrt{\boldsymbol{\mathrm{q}}}\:\right)^{\mathrm{2}} \:\geqslant\:\mathrm{0} \\ $$$$\:\:\boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}}\geqslant\:\sqrt{\mathrm{4}\boldsymbol{\mathrm{pq}}} \\ $$

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