Question

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68. If E= alog b-log c + blog c-log a + clog a-log b
where a, b, c > 0, then : Emin
(a) 0
(b) 1
(c) 3
(d) 9​

Answer 1

Correct Question :-

If a, b, c > 0, then find the minimum value of E,

$\bf \:E = {a}^{logb - logc} + {b}^{logc - loga} + {c}^{loga - logb}$

Answer

Given :-

$\bf \:E = {a}^{logb - logc} + {b}^{logc - loga} + {c}^{loga - logb}$

To find :-

• Minimum value of E.

Concept used :-

• AM  ≥ GM
• log1 = 0
• log(ab) = loga + logb

Solution :-

$\bf \:Let \: y = {a}^{logb - logc} \times {b}^{logc - loga} \times {c}^{loga - logb}$

Taking log on both sides, we get

$\bf \:logy =log( {a}^{logb - logc} \times {b}^{logc - loga} \times {c}^{loga - logb} )$

⟹ logy = (logb - logc) × loga + (logc - loga) × logb + (loga - logb) × logc

⟹ logy = logb × loga - logc × loga + logc × logb - loga × logb + loga × logb - logb × logc

⟹ logy = 0

⟹ y = 1

$\bf\implies \: {a}^{logb - logc} \times {b}^{logc - loga} \times {c}^{loga - logb} = 1$

Now, using AM  ≥  GM, we get

$\bf \:\dfrac{{a}^{logb - logc} + {b}^{logc - loga} + {c}^{loga - logb} }{3}  ≥  {( {a}^{logb - logc} \times {b}^{logc - loga} \times {c}^{loga - logb}) }^{\dfrac{1}{3} }$

$\bf \:\dfrac{{a}^{logb - logc} + {b}^{logc - loga} + {c}^{loga - logb} }{3}  ≥ {(1)}^{\dfrac{1}{3} }$

$\bf \:\dfrac{{a}^{logb - logc} + {b}^{logc - loga} + {c}^{loga - logb} }{3}  ≥1$

$\bf ⟹\: {a}^{logb - logc} + {b}^{logc - loga} + {c}^{loga - logb}  ≥ 3$

$\bf\implies \:minimum \: value \: of \: E =3$