Use Cauchy–Schwarz inequality to prove that for three positive reals $a, b, c$ such that $a + b + c \leqslant 3$ then $\frac{1}{\sqrt{a}} +\frac{1}{\sqrt{b}} +\frac{1}{\sqrt{c}} \geqslant 3$. (Use C-S for $\sqrt{a} , \sqrt{b} , \sqrt{c}$ and $\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c} }$).

How do I solve this?


From Cauchy Schwarz inequality we have that $\frac{1}{\sqrt{a}},\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}}$ and $\sqrt{a},\sqrt{b},\sqrt{c}$ Then $$(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq (\frac{1}{\sqrt{a}} \cdot \sqrt{a}+\frac{1}{\sqrt{b}} \cdot \sqrt{b}+\frac{1}{\sqrt{c}} \cdot \sqrt{c})^2=(1+1+1)^2=9$$ So $$(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 9$$ In another hand for any $x,y,z$ we have $$3(x^2+y^2+z^2)\geq (x+y+z)^2$$ because after expending you get that $$x^2+y^2+z^2\geq xy+yz+zx$$ So back to our question we have that for $x=\sqrt{a},y=\sqrt{b},z=\sqrt{c}$ $$3(a+b+c)\geq (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$$ since a+b+c=3 and getting ride of square then $$3\geq (\sqrt{a}+\sqrt{b}+\sqrt{c})$$ From first inequality we have that $$3 (\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})\geq (\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 9$$ or $$\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\geq 3$$

  • $\begingroup$ $√a+√b+√c=1·√a+1·√b+1·√c\le\sqrt{3}·\sqrt{a+b+c}$ again by Cauchy-Schwarz, no need to have long calculations. $\endgroup$ – LutzL Nov 13 '16 at 13:13
  • $\begingroup$ $x,y,z$ we have $$3(x^2+y^2+z^2)\geq (x+y+z)^2$$ the next question is to prove this how do I prove this? $\endgroup$ – user135643 Nov 13 '16 at 13:15
  • $\begingroup$ @arberavdullahu how do we get this $x,y,z$ we have $$3(x^2+y^2+z^2)\geq (x+y+z)^2$$ $\endgroup$ – user135643 Nov 13 '16 at 13:48
  • $\begingroup$ @LutzL how do we get this $x,y,z$ we have $$3(x^2+y^2+z^2)\geq (x+y+z)^2$$ $\endgroup$ – user135643 Nov 13 '16 at 13:51
  • $\begingroup$ @user135643: As already said twice, you can also use CSI for it: $$(1·x+1·y+1·z)^2\le(1^2+1^2+1^2)·(x^2+y^2+z^2).$$ $\endgroup$ – LutzL Nov 13 '16 at 14:14

This looks more like the identity between the harmonic and quadratic mean $$ \left(\frac{x_1^{-1}+x_2^{-1}+…+x_n^{-1}}n\right)^{-1}\le\left(\frac{x_1^2+x_2^2+…+x_n^2}n\right)^{\tfrac12} $$ with $n=3$, $x_1=\sqrt a$, $x_2=\sqrt b$, $x_3=\sqrt c$.

Or you can apply CSI twice $$ n=x_1\frac1{x_1}+x_2\frac1{x_2}+…+x_n\frac1{x_n}\le\sqrt{x_1^2+x_2^2+…+x_n^2}\sqrt{\frac1{x_1^2}+\frac1{x_2^2}+…+\frac1{x_n^2}} \\ \le\sqrt{\sqrt{n}\sqrt{x_1^4+x_2^4+…+x_n^4}}\sqrt{\frac1{x_1^2}+\frac1{x_2^2}+…+\frac1{x_n^2}} $$ with $x_1=\sqrt[4]a$ etc. This, after squaring and replacing $y_k=x_k^2$ gives again the harmonic-quadratic mean inequality.


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