# Correct the strategy about :$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$

inequalitysolution-verification

Problem :

Let $$a$$, $$b$$ and $$c$$ be positive numbers. Prove that:
$$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$

My strategy :

I introduce the function :

$$f\left(x\right)=\frac{x}{2+\frac{p^{2}\left(x+u\right)^{2}}{\left(x+k\right)^{2}}}$$

And :

$$b=\frac{a\left(a+u\right)p}{a+k},c=\frac{bp\left(b+u\right)}{b+k}$$

And :

Now the problem is :

$$f\left(a\right)+f\left(b\right)+\frac{c^{3}}{2c^{2}+a^{2}}-\frac{\left(a+b+c\right)}{3}\geq 0$$

Fact 1 :

It seems we can use directly Jensen's inequality as $$f(x)$$ is convex for $$u\geq k>0$$ and $$p>0$$ and $$x>0$$.

Now remains to show that it is positive or after a bit of algebra and using again Jensen's inequality .

$$g(a,u,k)=\frac{c^{3}}{2c^{2}+a^{2}}+2f\left(\frac{\left(a+b\right)}{2}\right)-\frac{\left(a+b+c\right)}{3}\geq 0$$
Wich is not hard again .

By not hard I mean using Buffalo's way all the coefficient are positives (I use Geogebra) using rational number wich is sufficient to show the strategy .

Replacing by the differents constraints we have and $$p=1$$:

$$g(x,y+x+z,y+x)=z² x (8x² + 3x z + 8x y + z y + 2y²) (6144x⁹ + 6528x⁸ z + 27648x⁸ y + 3712x⁷ z² + 24192x⁷ z y + 55296x⁷ y² + 1920x⁶ z³ + 11488x⁶ z² y + 39072x⁶ z y² + 64512x⁶ y³ + 678x⁵ z⁴ + 5244x⁵ z³ y + 15360x⁵ z² y² + 35904x⁵ z y³ + 48384x⁵ y⁴ + 99x⁴ z⁵ + 1549x⁴ z⁴ y + 6144x⁴ z³ y² + 11536x⁴ z² y³ + 20520x⁴ z y⁴ + 24192x⁴ y⁵ + 174x³ z⁵ y + 1461x³ z⁴ y² + 3939x³ z³ y³ + 5272x³ z² y⁴ + 7464x³ z y⁵ + 8064x³ y⁶ + 119x² z⁵ y² + 700x² z⁴ y³ + 1449x² z³ y⁴ + 1470x² z² y⁵ + 1686x² z y⁶ + 1728x² y⁷ + 36x z⁵ y³ + 168x z⁴ y⁴ + 288x z³ y⁵ + 232x z² y⁶ + 216x z y⁷ + 216x y⁸ + 4z⁵ y⁴ + 16z⁴ y⁵ + 24z³ y⁶ + 16z² y⁷ + 12z y⁸ + 12y⁹) / (3 (2x + y) (4x² + x z + 4x y + y²) (192x⁴ + 112x³ z + 384x³ y + 27x² z² + 144x² z y + 288x² y² + 20x z² y + 60x z y² + 96x y³ + 4z² y² + 8z y³ + 12y⁴) (192x⁶ + 352x⁵ z + 576x⁵ y + 300x⁴ z² + 832x⁴ z y + 720x⁴ y² + 120x³ z³ + 540x³ z² y + 784x³ z y² + 480x³ y³ + 18x² z⁴ + 148x² z³ y + 363x² z² y² + 368x² z y³ + 180x² y⁴ + 12x z⁴ y + 60x z³ y² + 108x z² y³ + 86x z y⁴ + 36x y⁵ + 2z⁴ y² + 8z³ y³ + 12z² y⁴ + 8z y⁵ + 3y⁶))$$

Where $$x=a$$ and $$k=x+y$$ and $$u=x+y+z$$. The same approach works with $$k\leq a \leq u$$ and $$k\leq u \leq a$$

Edit 08/05/2022 :

Using :

$$b=\frac{a\left(a+u\right)p}{a+k},c=\frac{bp\left(v+u\right)}{v+k}$$

And following the same path we have taking Buffalo's way in this case Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$ :

$$h(1,1+x+y+z,1+x+y,1+x)=(2304x¹³ z + 23616x¹² y z + 7056x¹² z² + 72576x¹² z + 107136x¹¹ y² z + 67560x¹¹ y z² + 652032x¹¹ y z + 6000x¹¹ z³ + 246000x¹¹ z² + 930816x¹¹ z + 284832x¹⁰ y³ z + 280356x¹⁰ y² z² + 2597184x¹⁰ y² z + 60992x¹⁰ y z³ + 1997856x¹⁰ y z² + 7417152x¹⁰ y z + 2704x¹⁰ z⁴ + 319888x¹⁰ z³ + 3172224x¹⁰ z² + 6724224x¹⁰ z + 494064x⁹ y⁴ z + 668334x⁹ y³ z² + 6051456x⁹ y³ z + 248132x⁹ y² z³ + 7097820x⁹ y² z² + 26078976x⁹ y² z + 32388x⁹ y z⁴ + 2394408x⁹ y z³ + 22505424x⁹ y z² + 47508480x⁹ y z + 1372x⁹ z⁵ + 269912x⁹ z⁴ + 4502176x⁹ z³ + 21966240x⁹ z² + 31207680x⁹ z + 587844x⁸ y⁵ z + 1017642x⁸ y⁴ z² + 9154008x⁸ y⁴ z + 551028x⁸ y³ z³ + 14523852x⁸ y³ z² + 53257536x⁸ y³ z + 129716x⁸ y² z⁴ + 7669964x⁸ y² z³ + 69551484x⁸ y² z² + 146876544x⁸ y² z + 14360x⁸ y z⁵ + 1812792x⁸ y z⁴ + 28509176x⁸ y z³ + 136484256x⁸ y z² + 194316480x⁸ y z + 1196x⁸ z⁶ + 167596x⁸ z⁵ + 4103560x⁸ z⁴ + 30746720x⁸ z³ + 94671120x⁸ z² + 99533952x⁸ z + 489960x⁷ y⁶ z + 1036836x⁷ y⁵ z² + 9422352x⁷ y⁵ z + 749550x⁷ y⁴ z³ + 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Where I only put the numerator and with the function $$h(a,u,k,v)$$

Question : Is my strategy correct ? If not can someone point out my mistakes ?