Does this "symmetric" inequality holds?Let σ1≤σ2,σ1⋅&lt;σ2⋅ be positive real numbers and suppose that σ1σ2=σ1⋅σ2⋅Is it true that1−(σ1+σ2)≥(σ1⋅)2+(σ2⋅)2−σ1⋅σ1−σ2⋅σ2,,,,?

σ1=σ2=13,σ1⋅=3−56 and σ2⋅=3+56But for σ1+σ2≥4σ1σ2 the reversed inequality is true.Indeed, let σ1=a, where b≥a&gt;0 and a+b≥4ab.Thus, σ1⋅ and σ2⋅ are roots of the equation:x2−x+ab=0,which givesσ1⋅=1−1−4ab2and σ2⋅=1+1−4ab2.Id est, we need to prove that:1−a−b+a⋅1−1−4ab2+b⋅1+1−4ab2≤1−2abor (b−a)1−4ab≤a+b−4abor (a−b)2(1−4ab)≤(a+b−4ab)2or ab(a+b−1)2≥0

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High School
Algebra
Algebra I
Quadratic function and equation

amonitas3zeb

Answered question

2022-04-02

Does this "symmetric" inequality holds?
Let

σ_{1} \leq σ_{2}, σ_{1}^{\cdot} < σ_{2}^{\cdot}

be positive real numbers and suppose that

σ_{1} σ_{2} = σ_{1}^{\cdot} σ_{2}^{\cdot}

Is it true that

1 - (σ_{1} + σ_{2}) \geq {(σ_{1}^{\cdot})}^{2} + {(σ_{2}^{\cdot})}^{2} - σ_{1}^{\cdot} σ_{1} - σ_{2}^{\cdot} σ_{2},,,, ?

Answer & Explanation

Korbin Rivera

Beginner2022-04-03Added 11 answers

σ_{1} = σ_{2} = \frac{1}{3}, σ_{1}^{\cdot} = \frac{3 - \sqrt{5}}{6}

and

σ_{2}^{\cdot} = \frac{3 + \sqrt{5}}{6}

But for

σ_{1} + σ_{2} \geq 4 σ_{1} σ_{2}

the reversed inequality is true.
Indeed, let

σ_{1} = a

, where

b \geq a > 0

and

a + b \geq 4 a b

.
Thus,

σ_{1}^{\cdot}

and

σ_{2}^{\cdot}

are roots of the equation:

x^{2} - x + a b = 0

,
which gives

σ_{1}^{\cdot} = \frac{1 - \sqrt{1 - 4 a b}}{2}

and

σ_{2}^{\cdot} = \frac{1 + \sqrt{1 - 4 a b}}{2} .

Id est, we need to prove that:

1 - a - b + a \cdot \frac{1 - \sqrt{1 - 4 a b}}{2} + b \cdot \frac{1 + \sqrt{1 - 4 a b}}{2} \leq 1 - 2 a b

(b - a) \sqrt{1 - 4 a b} \leq a + b - 4 a b

{(a - b)}^{2} (1 - 4 a b) \leq {(a + b - 4 a b)}^{2}

a b {(a + b - 1)}^{2} \geq 0

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