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Question-156164




Question Number 156164 by MathSh last updated on 08/Oct/21
Answered by ghimisi last updated on 09/Oct/21
∣x∣=(1/2)∣x+y+x−y∣≤(1/2)∣x+y∣+(1/2)∣y−x∣  (1)  ∣y∣=(1/2)∣y+x+y−x∣≤(1/2)∣x+y∣+(1/2)∣y−x∣∣  (2)  ∣3x+2y∣=∣4x+3y−(x+y)∣≤∣4x+3y∣+∣x+y∣  (3)  (1)+(2)+(3)⇒  ∣x∣+∣y∣+∣3x+2y∣≤∣4x+3y∣+2∣x+y∣+∣y−x∣
$$\mid{x}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mid{x}+{y}+{x}−{y}\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}}\mid{x}+{y}\mid+\frac{\mathrm{1}}{\mathrm{2}}\mid{y}−{x}\mid\:\:\left(\mathrm{1}\right) \\ $$$$\mid{y}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mid{y}+{x}+{y}−{x}\mid\leqslant\frac{\mathrm{1}}{\mathrm{2}}\mid{x}+{y}\mid+\frac{\mathrm{1}}{\mathrm{2}}\mid{y}−{x}\mid\mid\:\:\left(\mathrm{2}\right) \\ $$$$\mid\mathrm{3}{x}+\mathrm{2}{y}\mid=\mid\mathrm{4}{x}+\mathrm{3}{y}−\left({x}+{y}\right)\mid\leqslant\mid\mathrm{4}{x}+\mathrm{3}{y}\mid+\mid{x}+{y}\mid\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+\left(\mathrm{3}\right)\Rightarrow \\ $$$$\mid{x}\mid+\mid{y}\mid+\mid\mathrm{3}{x}+\mathrm{2}{y}\mid\leqslant\mid\mathrm{4}{x}+\mathrm{3}{y}\mid+\mathrm{2}\mid{x}+{y}\mid+\mid{y}−{x}\mid \\ $$
Commented by MathSh last updated on 09/Oct/21
Perfect dear Ser, thank you
$$\mathrm{Perfect}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you} \\ $$

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