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

Prove in detail the inequality of this formula:
Let x, y, z positive real number and $\mathrm{△}ABS$ a triangle. $\left[ABC\right]$ denotes the triangle area and a, b, c the sides of the triangle. The inequality below is true:
${a}^{2}x+{b}^{2}y+{c}^{2}z\ge 4\left[ABC\right]\sqrt{xy+xz+yz}$
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stamptsk

Step 1
Proof
Let $\alpha ,\beta ,\gamma$ denote the opposite angles to the sides $a,b,c$ respectively. R is the circumradius of $\mathrm{△}ABC$. Observe that:
${a}^{2}x+{b}^{2}y+{c}^{2}z\ge 4\left[ABC\right]\sqrt{xy+xz+yz}$
${a}^{2}x+{b}^{2}y+{c}^{2}Z\ge \frac{abc}{R}\sqrt{xy+xz+yz}$
$\frac{aRx}{bc}+\frac{bRy}{ac}+\frac{cRz}{ab}\ge \sqrt{xy+xz+yz}$
$\frac{1}{2}\left(\frac{4a{R}^{2}x}{2Rbc}+\frac{4b{R}^{2}y}{2Rac}+\frac{4c{R}^{2}z}{2Rab}\right)\ge \sqrt{xy+xz+yz}$
$x\frac{\mathrm{sin}\alpha }{\mathrm{sin}\beta \mathrm{sin}\gamma }+y\frac{\mathrm{sin}\beta }{\mathrm{sin}\alpha \mathrm{sin}\gamma }+z\frac{\mathrm{sin}\gamma }{\mathrm{sin}\alpha \mathrm{sin}\beta }\ge 2\sqrt{xy+xz+yz}$
$x\frac{\mathrm{sin}\left(\pi -\alpha \right)}{\mathrm{sin}\beta \mathrm{sin}\gamma }+y\frac{\mathrm{sin}\left(\pi -\beta \right)}{\mathrm{sin}\alpha \mathrm{sin}\gamma }+z\frac{\mathrm{sin}\left(\pi -\gamma \right)}{\mathrm{sin}\alpha \mathrm{sin}\beta }\ge 2\sqrt{xy+xz+yz}$
$x\frac{\mathrm{sin}\left(\alpha +\beta +\gamma -\alpha \right)}{\mathrm{sin}\beta \mathrm{sin}\gamma }+y\frac{\mathrm{sin}\left(\alpha +\beta +\gamma -\beta \right)}{\mathrm{sin}\alpha \mathrm{sin}\gamma }+z\frac{\alpha +\beta +\gamma -\gamma }{\right\}}\left\{\mathrm{sin}\alpha \mathrm{sin}\beta \right\}\ge 2\sqrt{xy+xz+yz}$

###### Not exactly what you’re looking for?
Eliza Norris
Step 1
Here is my algebraic proof.
We need to prove that:
${\left({a}^{2}x+{b}^{2}y+{c}^{2}z\right)}^{2}\ge \sum _{cyc}\left(2{a}^{2}{b}^{2}-{a}^{4}\right)\left(xy+xz+yz\right)$
or
${c}^{4}{z}^{2}-\left(\left(\sum _{cyc}\left(2{a}^{2}{b}^{2}-{a}^{4}\right)-2{a}^{2}{c}^{2}\right)x+\left(\sum _{cyc}\left(2{a}^{2}{b}^{2}{a}^{4}\right)-2{b}^{2}{c}^{2}\right)y\right)z+$
$+{a}^{4}{x}^{2}+{b}^{4}{y}^{2}-\left(\sum _{cyc}\left(2{a}^{2}{b}^{2}-{a}^{4}\right)-2{a}^{2}{b}^{2}\right)xy\ge 0$
for which its
###### Not exactly what you’re looking for?
RizerMix
Step 1 Writing the expression as $\frac{{a}^{2}x+{b}^{2}y+{c}^{2}z}{\sqrt{xy+yz+zx}}\ge 4\left[ABC\right]$ 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 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 by AM-GM, and clearly $2ab\ge 2ab\mathrm{sin}\gamma =4\left[ABC\right]$ The last case is $x,y,z>0$. By the Lagrange method we obtain a critical point in the interior where that is $x={b}^{2}+{c}^{2}-{a}^{2}$ $y={c}^{2}+{a}^{2}-{b}^{2}$ $z={a}^{2}+{b}^{2}-{c}^{2}$ up to a multiplicative constant. Substituting above we obtain ${a}^{2}x+{b}^{2}y+{c}^{2}z=2\left({a}^{2}{b}^{2}+{b}^{2}{c}^{2}+{c}^{2}{a}^{2}\right)-\left({a}^{4}+{b}^{4}+{c}^{4}\right)=16\left[ABC{\right]}^{2}$ by Herons