Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $

Question

Question -

Let $a, b, c$ be positive real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ . Prove that $$ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $$

My try

$$ \frac{a^{2}}{a+2 b^{2}}=a-\frac{2 a b^{2}}{a+2 b^{2}} \geq a-\frac{2 a b^{2}}{3 \sqrt[3]{a b^{4}}}=a-\frac{2(a b)^{2 / 3}}{3} $$ which implies that $$ \sum_{c y c} \frac{a^{2}}{a+2 b^{2}} \geq \sum_{c y c} a-\frac{2}{3} \sum_{c y c}(a b)^{\frac{2}{3}} $$ It suffices to prove that $$ (a b)^{2 / 3}+( b c)^{2 / 3}+\left (c a)^{2 / 3} \leq 3\right. $$

beacuse we can easily get that $\sum a \ge 3$

but i am not able to prove it..

note that we have to prove this using only am-gm or weighted am-gm or power mean or any kind of means inequality because author did not introduce any advance inequality yet...

any hints ???

thankyou

Answer 1

It is enough to show that $\sqrt a + \sqrt b + \sqrt c = 3 \implies (ab)^{2/3} + (bc)^{2/3} + (ca)^{2/3} \leqslant 3$. For ease, let us replace $a, b, c$ with $x^6, y^6, z^6$, so we need to show for positives, $x^3+y^3+z^3=3 \implies (xy)^4+(yz)^4+(zx)^4\leqslant 3$. This one is in fact a known old chestnut.

We note by AM-GM, $xy \leqslant \frac13(x^3+y^3+1) = \frac13(4-z^3)$, hence $(xy)^4 \leqslant \frac13(4x^3y^3-x^3y^3z^3)$. Cyclically summing three such inequalities, we get $$\sum (xy)^4 \leqslant \frac43\sum (xy)^3-(xyz)^3$$ Now with $X=x^3, Y = y^3, Z = z^3$, it is enough to show with $X+Y+Z=3$ $$4(XY+YZ+ZX) -3XYZ\leqslant 9$$ which is the well known Schur's inequality: $$4(X+Y+Z)(XY+YZ+ZX) \leqslant (X+Y+Z)^3+9XYZ$$

Answer 2

Elaborating on the above solution:

We first wish to prove that the last inequality holds. Schur's inequality in its simplest form states that: For non-negative real numbers $x,y,z$, we have $x^3+y^3+z^3+3xyz \geq xy(x+y)+xz(x+z)+yz(y+z)$.

Now, expanding the R.H.S., we obtain: $(X+Y+Z)^3+9XYZ$

$=X^3+3X^2Y+3X^2Z+3XY^2+6XYZ+3XZ^2+Y^3+3Y^2Z+3YZ^2+Z^3+9XYZ$

$=X^3+Y^3+Z^3+3XYZ+12XYZ+3X^2Y+3X^2Z+3XY^2+3XZ^2+3Y^2Z+3YZ^2$

$\geq XY(X+Y) +XZ(X+Z) + YZ(Y+Z)+12XYZ+3X^2Y+3X^2Z+3XY^2+3XZ^2+3Y^2Z+3YZ^2$,

$=X^2Y+XY^2+X^2Z+XZ^2+Y^2Z+YZ^2+12XYZ+3X^2Y+3X^2Z+3XY^2+3XZ^2+3Y^2Z+3YZ^2$

$=4X^2Y+4XY^2+4X^2Z+4XZ^2+4Y^2Z+4YZ^2+12XYZ$

$=4(X^2Y+XYZ+XY^2+X^2Z+XYZ+XZ^2+Y^2Z+XYZ+YZ^2)$

$=4[XY(X+Y+Z)+XZ(X+Y+Z)+YZ(X+Y+Z)]$

$=4(X+Y+Z)(XY+YZ+ZX)$

Simply flipping the inequality sign gives us: $4(X+Y+Z)(XY+YZ+ZX) \leq (X+Y+Z)^3+9XYZ$, so we are done.

Finally, we observe that this inequality implies the second last inequality. With $X+Y+Z=3$, we have:

$4(X+Y+Z)(XY+YZ+ZX) \leq (X+Y+Z)^3+9XYZ$

$\Rightarrow 4(XY+YZ+ZX) \leq \ (X+Y+Z)^2+3XYZ$

$\Rightarrow 4(XY+YZ+ZX) -3XYZ \leq \ (X+Y+Z)^2 = 3^2 =9$

(Q.E.D.)

Answer 3

Another way.

By C-S and P-M we obtain: $$\sum_{cyc}\frac{a^2}{a+2b^2}-1=\sum_{cyc}\frac{a^4}{a^3+2a^2b^2}-1\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^3+2a^2b^2)}-1=$$ $$=\frac{\sum\limits_{cyc}(a^4-a^3)}{\sum\limits_{cyc}(a^3+2a^2b^2)}=\frac{9(a^4+b^4+c^4)-(a^3+b^3+c^3)(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{9\sum\limits_{cyc}(a^3+2a^2b^2)}\geq0.$$ Because by P-M $$\sqrt[4]{\frac{a^4+b^4+c^4}{3}}\geq\sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$ or $$\sqrt[4]{\left(\frac{a^4+b^4+c^4}{3}\right)^3}\geq\frac{a^3+b^3+c^3}{3}$$ or $$\sqrt[4]{3(a^4+b^4+c^4)^3}\geq a^3+b^3+c^3$$ and $$\sqrt[4]{\frac{a^4+b^4+c^4}{3}}\geq\left(\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}\right)^2$$ or $$\sqrt[4]{2187(a^4+b^4+c^4)}\geq(\sqrt a+\sqrt b+\sqrt c)^2.$$ Id est, $$9(a^4+b^4+c^4)=\sqrt[4]{3(a^4+b^4+c^4)^3}\cdot\sqrt[4]{2187(a^4+b^4+c^4)}\geq$$ $$\geq(a^3+b^3+c^3)(\sqrt{a}+\sqrt{b}+\sqrt{c})^2.$$