Prove in detail the inequality of this formula: Let x, y,

veceriraby

veceriraby

Answered question

2022-01-21

Prove in detail the inequality of this formula:
Let x, y, z positive real number and ABS a triangle. [ABC] denotes the triangle area and a, b, c the sides of the triangle. The inequality below is true:
a2x+b2y+c2z4[ABC]xy+xz+yz

Answer & Explanation

stamptsk

stamptsk

Beginner2022-01-22Added 23 answers

Step 1
Proof
Let α,β,γ denote the opposite angles to the sides a,b,c respectively. R is the circumradius of ABC. Observe that:
a2x+b2y+c2z4[ABC]xy+xz+yz
a2x+b2y+c2ZabcRxy+xz+yz
aRxbc+bRyac+cRzabxy+xz+yz
12(4aR2x2Rbc+4bR2y2Rac+4cR2z2Rab)xy+xz+yz
xsinαsinβsinγ+ysinβsinαsinγ+zsinγsinαsinβ2xy+xz+yz
xsin(πα)sinβsinγ+ysin(πβ)sinαsinγ+zsin(πγ)sinαsinβ2xy+xz+yz
xsin(α+β+γα)sinβsinγ+ysin(α+β+γβ)sinαsinγ+zα+β+γγ}{sinαsinβ}2xy+xz+yz

Eliza Norris

Eliza Norris

Beginner2022-01-23Added 15 answers

Step 1
Here is my algebraic proof.
We need to prove that:
(a2x+b2y+c2z)2cyc(2a2b2a4)(xy+xz+yz)
or
c4z2((cyc(2a2b2a4)2a2c2)x+(cyc(2a2b2a4)2b2c2)y)z+
+a4x2+b4y2(cyc(2a2b2a4)2a2b2)xy0
for which its
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

Step 1 Writing the expression as a2x+b2y+c2zxy+yz+zx4[ABC] the RHS is independent of x,y,z, therefore it is necessary and sufficient to prove the inequality in the worst possible case, i.e. when the LHS is minimized in x,y,z. In particular, by homogeneity we can fix a,b,c and consider the problem min{a2x+b2y+c2z:xy+yz+zx=1, x, y, z0} If two of x,y,z are zero the inequality is trivially proven. If just one of them is zero, say z, then the problem becomes min{a2x+b2y:xy=1, x, y0}=2ab by AM-GM, and clearly 2ab2absinγ=4[ABC] The last case is x,y,z>0. By the Lagrange method we obtain a critical point in the interior where (a2, b2, c2)=λ(y+z, z+x, x+y) that is x=b2+c2a2 y=c2+a2b2 z=a2+b2c2 up to a multiplicative constant. Substituting above we obtain a2x+b2y+c2z=2(a2b2+b2c2+c2a2)(a4+b4+c4)=16[ABC]2 by Herons

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