Does this "symmetric" inequality holds? Let \(\displaystyle\sigma_{{1}}\le\sigma_{{2}},{\sigma_{{1}}^{\cdot}}{

amonitas3zeb

amonitas3zeb

Answered question

2022-04-02

Does this "symmetric" inequality holds?
Let σ1σ2,σ1<σ2 be positive real numbers and suppose that σ1σ2=σ1σ2
Is it true that
1(σ1+σ2)(σ1)2+(σ2)2σ1σ1σ2σ2,,,,?

Answer & Explanation

Korbin Rivera

Korbin Rivera

Beginner2022-04-03Added 11 answers

σ1=σ2=13,σ1=356 and σ2=3+56
But for σ1+σ24σ1σ2 the reversed inequality is true.
Indeed, let σ1=a, where ba>0 and a+b4ab.
Thus, σ1 and σ2 are roots of the equation:
x2x+ab=0,
which gives
σ1=114ab2
and σ2=1+14ab2.
Id est, we need to prove that:
1ab+a114ab2+b1+14ab212ab
or (ba)14aba+b4ab
or (ab)2(14ab)(a+b4ab)2
or ab(a+b1)20

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