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

veceriraby 2022-01-21 Answered
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
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Expert Answer

stamptsk
Answered 2022-01-22 Author has 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

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Eliza Norris
Answered 2022-01-23 Author has 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
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RizerMix
Answered 2022-01-27 Author has 438 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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