Inequality 2

[MM]

If [tex]a,b,c\ge 0[/tex] and [tex]a+b+c=\frac{3}{2}[/tex] prove that:
[tex]4\left(a^{3}+b^{3}+c^{3}\right)+12abc+9\ge 8\left(a^{2}\sqrt{b+c}+b^{2}\sqrt{c+a}+c^{2}\sqrt{a+b}+ab+bc+ca\right)[/tex]

I am the author.

[martosss]

Let us add [tex]4(a^2+b^2+c^2)[/tex] to both sides of the inequality:

[tex]4(a^3+b^3+c^3)+12abc+4(a^2+b^2+c^2)+N 9\ge 8(a^2\sqrt{b+c}+b^2\sqrt{a+c}+c^2\sqrt{a+b})+N {\underbrace{4(a+b+c)^2}_{=9}}[/tex]

Now from AM≥GM we have

[tex]\frac{(b+c)+1}{2}\ge\sqrt{b+c}[/tex]

[tex]\frac{(a+c)+1}{2}\ge\sqrt{a+c}[/tex]

[tex]\frac{(a+b)+1}{2}\ge\sqrt{a+b}[/tex]

Now let us multiply (1), (2) and (3) respectively with a², b² and c², then add the 3 inequalities and multiply that by 8 - we get

[tex]4\left[a^2(b+c)+b^2(a+c)+c^2(a+b)+a^2+b^2+c^2\right]\ge 8(a^2\sqrt{b+c}+b^2\sqrt{a+c}+c^2\sqrt{a+b})[/tex]

So we must prove, that

[tex]N 4(a^3+b^3+c^3)+\overbrace{N {12}}^{3}abc+N{4(a^2+b^2+c^2)}\ge N 4\left[a^2(b+c)+b^2(a+c)+c^2(a+b)\right]+N {4\left[a^2+b^2+c^2\right]}[/tex]

That leaves the Schur's Inequality and we're done

[billhood]

Now let us multiply (1), (2) and (3) respectively with a², b² and c², then add the 3 inequalities and multiply that by 8 - we get

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Nawaz