Solution pour le challenge 81
Proposons deux méthodes.
Solution 1
L’inégalité de Cauchy-Schwarz (un cas particulier de celle-ci, en fait) dit que pour tout quadruplet de réels :
![Rendered by QuickLaTeX.com u_{1}=u_{2}=1,](https://math-os.com/wp-content/ql-cache/quicklatex.com-405c8244db5a54b989ff1a1835e1514d_l3.png)
![Rendered by QuickLaTeX.com v_{1}=a\sqrt{b}](https://math-os.com/wp-content/ql-cache/quicklatex.com-5aafe1243a263196c3e767e9163a1c8d_l3.png)
![Rendered by QuickLaTeX.com v_{2}=b\sqrt{a},](https://math-os.com/wp-content/ql-cache/quicklatex.com-8e43506282b0ceb7ef819fbdeb6b2a47_l3.png)
![Rendered by QuickLaTeX.com 2ab\left(a+b\right)](https://math-os.com/wp-content/ql-cache/quicklatex.com-4556d8339c1788bc33720908785f5976_l3.png)
![Rendered by QuickLaTeX.com {\displaystyle \left[\frac{\left(a+b\right)^{2}}{2}+\frac{a+b}{4}\right]^{2}},](https://math-os.com/wp-content/ql-cache/quicklatex.com-49be4115cbb419a0fa4ec3e7ff77188a_l3.png)
![Rendered by QuickLaTeX.com 2ab](https://math-os.com/wp-content/ql-cache/quicklatex.com-cb780ad74b78a297028149d725632172_l3.png)
![Rendered by QuickLaTeX.com {\displaystyle \left(a+b\right)\left[\frac{a+b}{2}+\frac{1}{4}\right]^{2}}](https://math-os.com/wp-content/ql-cache/quicklatex.com-24a2c16e44b24c9227de2c64006e90ec_l3.png)
![Rendered by QuickLaTeX.com 32ab](https://math-os.com/wp-content/ql-cache/quicklatex.com-f961c9882b06a44b4c3f36cedb176805_l3.png)
![Rendered by QuickLaTeX.com \left(a+b\right)\left[2\left(a+b\right)+1\right]^{2}.](https://math-os.com/wp-content/ql-cache/quicklatex.com-e8df4193d1e7b751abeeaeb318db7981_l3.png)
![Rendered by QuickLaTeX.com a^{2}-6ab+b^{2}=\left(a-b\right)^{2}-4ab](https://math-os.com/wp-content/ql-cache/quicklatex.com-75b56160198f8137689aefcba55d434e_l3.png)
![Rendered by QuickLaTeX.com 4ab\leqslant\left(a+b\right)^{2};](https://math-os.com/wp-content/ql-cache/quicklatex.com-90a1750c6376dff19f978cf8223e9364_l3.png)
![Rendered by QuickLaTeX.com x=a+b\geqslant0](https://math-os.com/wp-content/ql-cache/quicklatex.com-4e87665286662a80997dba7bcfed3923_l3.png)
![Rendered by QuickLaTeX.com \delta\geqslant0,](https://math-os.com/wp-content/ql-cache/quicklatex.com-dc32934f00ef0d11d93ca3b1ceedad51_l3.png)
Solution 2
On passe en coordonnées polaires ! Si l’on pose et
avec
et
il s’agit de montrer que :
Pour consulter l’énoncé, c’est ici