Question -

Prove that for all non-negative real numbers a,b, c, we have $$ \sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2} $$

My work -

we may assume that $a b c=1$ The problem becomes $$ \sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}} \geq 2 \sqrt{2} $$ where $x=a^{3}, y=b^{3}, z=c^{3}$

now i did not know where to go from here ...i tried all classic inequalities like chebyshev,re-arangement but none of them working .

can anyone solve this using classic inequalities

any help will be appreciated

thank you


By C-S twice we obtain: $$\sum_{cyc}\sqrt{\frac{2a^2+bc}{a^2+2bc}}-2\sqrt2=\sum_{cyc}\frac{\sqrt{(2a^2+bc)(a^2+2bc)}}{a^2+2bc}-2\sqrt2\geq$$ $$\geq\sum_{cyc}\frac{\sqrt2(a^2+bc)}{a^2+2bc}-2\sqrt2=\sqrt2\left(\sum_{cyc}\left(\frac{a^2+bc}{a^2+2bc}-\frac{1}{2}\right)-\frac{1}{2}\right)=$$ $$=\sqrt2\left(\sum_{cyc}\frac{a^2}{2(a^2+2bc)}-\frac{1}{2}\right)\geq \sqrt2\left(\frac{(a+b+c)^2}{2\sum\limits_{cyc}(a^2+2bc)}-\frac{1}{2}\right)=0.$$

  • 1
    $\begingroup$ unexpected!! i think CS will not work here thats i did not even try it , but this is amazingly simple proof.thanks $\endgroup$ – Ishan May 5 '20 at 10:46
  • $\begingroup$ @Ishan You are welcome! $\endgroup$ – Michael Rozenberg May 5 '20 at 10:58

Take $$ \sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}} $$ where $x=\frac{a^2}{bc}>0, y=\frac{b^2}{ac}>0, z=\frac{c^2}{ab}>0$.

Note that functions $$f(w)=\sqrt{\frac{2 w+1}{w+2}}$$ are strictly increasing for $w\in[0,\infty)$.

Without the loss of generality assume that $x\geq y\geq z$. Hence $$\frac{a^2}{bc}\geq \frac{b^2}{ac}\geq \frac{c^2}{ab},\, a\geq b\geq c.$$ Furthermore $$x\geq\frac{a^2}{c^2}=e^2,\,y\geq\frac{c^2}{a^2}=\frac{1}{e^2},\,z\geq\frac{c^2}{a^2}=\frac{1}{e^2}.$$ Using all of this we get $$ \sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}} \geq \sqrt{\frac{2 e^2+1}{e^2+2}}+2\sqrt{\frac{2+e^2}{1+2e^2}}. $$ Note that function on the right hand side has an infimum $$\inf\limits_{e\rightarrow\infty}\sqrt{\frac{2 e^2+1}{e^2+2}}+2\sqrt{\frac{2+e^2}{1+2e^2}}=2\sqrt{2}$$ So, we get $$ \sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}}\geq 2\sqrt{2}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.